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## Triangles Assignments Class 9 Mathematics

**Solved Examples**

**Example : Observe the following figure. **

**Ajay wishes to determine the distance between two objects A and B, but there is a house in between. So, he devises an ingenious way to fix the problem. First, he fixes a pole at any point O so that both A and B are visible from O. He then fixes another pole at point D which is collinear to point O and object A, and is at the same distance from O as A, i.e., DO = AO.****Similarly, he fixes a pole at point C which is collinear to point O and object B, and is at the same distance from O as B, i.e., CO = BO. Finally, he measures CD to find the distance between A and B. How can Ajay be sure that CD = AB?****Solution:**

We have two triangles in the given figure, i.e., Î”AOB and Î”DOC.

In these two triangles, we have:

AO = DO (Given)

âˆ AOB = âˆ DOC (Vertically opposite angles)

BO = CO (Given)

Therefore, by the SAS congruence rule, we can say that:

Î”AOB â‰… Î”DOC

â‡’ AB = CD (By CPCT)

This is the reason why Ajay measures CD to find the distance between objects A and B.

**Example 2: In the given quadrilateral PQRS, PR bisects âˆ QPS and PQ = PS. Prove that:****i) Î”PQR â‰…Î”PSR****ii) QR = SR **

**Solution:**

i) In Î”PQR and Î”PSR, we have:

PQ = PS(Given)

PR = PR(Common side)

âˆ QPR = âˆ SPR(&because PR bisects âˆ QPS)

So, by the SAS congruence rule, we obtain:

Î”PQR â‰… Î”PSR

ii) We have proved that Î”PQR â‰… Î”PSR.

âˆ´ QR = SR (&because Corresponding parts of congruent triangles are equal)

**Example 1: In the shown figure, PR = QS and âˆ QPR = âˆ PQS. Prove that Î”PQR â‰… Î”QPS. Also, show that PS = QR and âˆ QPS = âˆ PQR. **

**Solution:**

In Î”PQR and Î”QPS, we have:

PR = QS(Given)

âˆ QPR = âˆ PQS(Given)

PQ = PQ(Common side)

âˆ´ Î”PQR â‰… Î”QPS(By the SAS congruence criterion)

â‡’ PS = QR and âˆ QPS = âˆ PQR (By CPCT)

**Example 2: Prove that Î”PQR is isosceles if the altitude drawn from a vertex bisects the opposite side.****Solution:**

The given figure shows the Î”PQR having PS as an altitude that bisects the opposite side QR.

In Î”PSQ and Î”PSR, we have:

QS = SR (&because Altitude PS bisects QR)

PS = PS(Common side)

âˆ PSQ = âˆ PSR = 90Â°(&because PS is the altitude to QR)

âˆ´ Î”PSQ â‰… Î”PSR(By the SAS congruence rule)

â‡’ PQ = PR(By CPCT)

Therefore, Î”PQR is isosceles.

**Example 3: If the angle bisector of any angle of a triangle bisects the opposite side then show that the triangle is isosceles.****Solution:**

Let Î”PQR be the given triangle and PS is the angle bisector of âˆ QPR such that it bisects the side QR.

Let us extend the segment PS to point T such that PS = TS.

In Î”PQS and Î”TRS, we have

QS = RS (Given)

âˆ PSQ = âˆ TSR (Vertically opposite angles)

PS = TS (By construction)

So, by the SAS congruence criterion, we have:

Î”PQS â‰… Î”TRS

By CPCT, we obtain

PQ = TR …(1)

And âˆ QPS = âˆ RTS …(2)

But âˆ QPS = âˆ RPS …(3) (PS bisects âˆ QPR)

âˆ´ âˆ RTS = âˆ RPS [From (2) and (3)]

â‡’ PR = TR (Sides opposite to equal angles)

âˆ´ PQ = PR [From (1)]

Thus, Î”PQR is an isosceles triangle.

**Example 1: ABCD is a quadrilateral in which AD = BC and âˆ DAB = âˆ CBA.**

**Prove that:****i) Î”ABD â‰… Î”BAC****ii) BD = AC****iii) âˆ ABD = âˆ BAC****Solution:**

i) In Î”ABD and Î”BAC, we have:

AD = BC (Given)

âˆ DAB = âˆ CBA (Given)

AB = BA (Common side)

So, by the SAS congruence criterion, we have:

Î”ABD â‰… Î”BAC

ii) We have proved that Î”ABD â‰… Î”BAC.

âˆ´ BD = AC (By CPCT)

iii) Since Î”ABD â‰… Î”BAC, we have:

âˆ ABD = âˆ BAC(By CPCT)

**Example 2: In the given figure, PR = PO, PQ = PS and âˆ QPS = âˆ OPR. Show that QR = SO. **

**Solution:**

It is given that âˆ QPS = âˆ OPR.

âˆ´ âˆ QPS + âˆ SPR = âˆ OPR + âˆ SPR

â‡’ âˆ QPR = âˆ SPOâ€¦ (1)

In Î”QPR and Î”SPO, we have:

PQ = PS (Given)

âˆ QPR = âˆ SPO (From equation 1)

PR = PO (Given)

So, by the SAS congruence rule, we have:

Î”QPR â‰… Î”SPO

â‡’ QR = SO (By CPCT)

**Example 3: In an isosceles triangle, prove that the medians on the equal sides are equal.****Solution:**

Let Î”PQR be an isosceles triangle such that PQ = PR.

Also, let RS and QT be the medians to the sides PQ and PR respectively.

In Î”PQR, we have

PS = SQ = 1/2 PQ (RS is the median)

And PT = TR = 1/2 PR (QT is the median)

But PQ = PR

âˆ´ PS = SQ = PT = TR …(1)

In Î”PRS and Î”PQT, we have

PQ = PR (Given)

âˆ RPS = âˆ QPT (Common angle)

PS = PT [From (1)]

So, by the SAS congruence rule, we have:

Î”PRS â‰… Î”PQT

âˆ´ RS = QT (By CPCT)

Thus, the medians on the equal sides of an isosceles triangle are equal.

Proving Theorem of Right Angled Triangle

There is a theorem of right angled triangles which states that:

**Example 1: Check whether the given triangles are congruent or not. **

**Solution:**

In Î”ABC, we have:

âˆ ABC + âˆ BCA + âˆ BAC = 180Â° (By the angle sum property)

â‡’ 50Â° + 60Â° + âˆ BAC = 180Â°

â‡’ 110Â° + âˆ BAC = 180Â°

â‡’ âˆ BAC = 180Â° âˆ’ 110Â°

â‡’ âˆ BAC = 70Â°

In Î”ABC and Î”DEF, we have:

âˆ BAC = âˆ EDF = 70Â°

AB = DE (Given)

âˆ ABC = âˆ DEF = 50Â°

Therefore, by the ASA congruence rule, we have:

Î”ABC â‰… Î”DEF

**Example 2: In the given figure, AB and CD are two equal and parallel lines. Prove that Î”ABO â‰… Î”CDO. **

**Solution:**

It is given that AB||CD. AD and BC are transversals lying on lines AB and CD.

So, by the alternate angles axiom, we obtain:

âˆ OAB = âˆ ODCâ€¦ (1)

âˆ OBA = âˆ OCDâ€¦ (2)

In Î”ABO and Î”CDO, we have:

âˆ OAB = âˆ ODC (By equation 1)

AB = CD(Given)

âˆ OBA = âˆ OCD (By equation 2)

Thus, by the ASA congruence rule, we obtain:

Î”ABO â‰… Î”CDO

**Example 1: In the given quadrilateral ABCD, diagonal AC bisects âˆ BAD and âˆ BCD. Prove that AB = AD and CB = CD. **

**Solution:**

Since diagonal AC bisects âˆ BAD and âˆ CAD, we have:

âˆ BAC = âˆ DAC and âˆ BCA = âˆ DCA

In Î”ACB and Î”ACD, we have:

âˆ BAC = âˆ DAC (Given)

âˆ BCA = âˆ DCA (Given)

AC = AC (Common side)

So, by the ASA congruence rule, we have:

Î”ACB â‰… Î”ACD

â‡’ AB = AD and CB = CD (By CPCT)

**Example 2: Consider the two triangular parks ABC and DEF shown below. **

**Tina jogs around park ABC and Aliya jogs around park DEF daily. Paths AB and DE are equal in length. If both girls jog an equal number of rounds daily, then check whether or not they cover the same distance while jogging? ****Solution:**

In Î”ABC and Î”DEF, we have:

âˆ BAC = âˆ EDF = 70Â°(Given)

AB = DE(Given)

âˆ ABC = âˆ DEF = 50Â°(Given)

Therefore, by the ASA congruency rule, we obtain:

Î”ABC â‰… Î”DEF

â‡’ AC = DF and BC = EF(By CPCT)

âˆ´ AB + BC + CA = DE + EF + FD

Hence, both Tina and Aliya cover the same distance daily while jogging.

**Example 1: The given Î”ABC is isosceles with AB = AC. âˆ LOC = 2âˆ OBC and âˆ MOB = 2âˆ OCB.** **Prove that Î”BCM â‰… Î”CBL. **

**Solution:**

It is given that:

âˆ LOC = 2âˆ OBCâ€¦ (1)

âˆ MOB = 2âˆ OCBâ€¦ (2)

Now, âˆ LOC = âˆ MOB (Vertically opposite angles)

Using equations (1) and (2), we obtain:

âˆ OCB = âˆ OBC

â‡’ âˆ MCB = âˆ LBCâ€¦ (3)

In Î”BCM and Î”CBL, we have:

âˆ MBC = âˆ LCB (âˆµ Î”ABC is isosceles with AB = AC)

BC = CB (Common side)

âˆ MCB = âˆ LBC (Using equation 3)

Thus, by the ASA congruence rule, we obtain:

Î”BCM â‰… Î”CBL

**Example 2: In the given figure, âˆ BCD = âˆ ADC and âˆ ACB = âˆ BDA. Prove that AD = BC and âˆ CAD = âˆ DBC. **

**Solution:**

It is given that:

âˆ BCD = âˆ ADC â€¦ (1)

âˆ ACB = âˆ BDA â€¦ (2)

On adding equations (1) and (2), we get:

âˆ BCD + âˆ ACB = âˆ ADC + âˆ BDA

â‡’ âˆ ACD = âˆ BDCâ€¦ (3)

In Î”ACD and Î”BDC, we have:

âˆ ADC = âˆ BCD(Given)

CD = DC (Common side)

âˆ ACD = âˆ BDC (By equation 3)

So, by the ASA congruence rule, we have:

Î”ACD â‰… Î”BDC

â‡’ AD = BC and âˆ CAD = âˆ DBC (By CPCT)

AAS Congruence Rule

**Example 1: The given Î”ABC is isosceles with AB = AC and âˆ ADC = âˆ AEB. Prove that Î”ABE â‰… Î”ACD. **

**Solution:**

In Î”ABE and Î”ACD, we have:

AB = AC (Given)

âˆ AEB = âˆ ADC (Given)

âˆ BAE = âˆ CAD (Common angle)

Thus, by the AAS congruence rule, we obtain:

Î”ABE â‰… Î”ACD

**Example 2: In the given figure, OR bisects âˆ POQ and A is any point on OR. AB and AC are the perpendiculars drawn from A to the arms OP and OQ respectively. Prove that Î”AOB â‰… Î”AOC. **

**Solution:**

It is given that OR bisects âˆ POQ.

âˆ´ âˆ POR = âˆ QOR

â‡’ âˆ BOA = âˆ COAâ€¦ (1)

In Î”AOB and Î”AOC, we have:

âˆ ABO = âˆ ACO = 90Â° (âˆµ AB and AC are perpendiculars)

âˆ BOA = âˆ COA (By equation 1)

AO = AO (Common side)

Thus, by the AAS congruence rule, we obtain:

Î”AOB â‰… Î”AOC

**Example 1: If two angles of a triangle are equal, then prove that the sides opposite them are also equal.****Solution:**

Consider a Î”PQR with âˆ PQR = âˆ PRQ.

We have to prove that PQ = PR.

Construction: Draw the bisector of âˆ QPR and let it meet side QR at point S.

In Î”PSQ and Î”PSR, we have:

âˆ PQS = âˆ PRS (Given)

PS = PS (Common side)

âˆ QPS = âˆ RPS (âˆµ PS bisects âˆ QPR)

So, by the AAS congruence rule, we obtain:

Î”PSQ â‰… Î”PSR

â‡’ PQ = PR (By CPCT)

Hence, we have proved that the sides opposite equal angles of a triangle are also equal.

**Example 2: DA and CB are equal perpendiculars to a line segment AB. Show that line segment CD bisects AB at point O. **

**Solution:**

In Î”DAO and Î”CBO, we have:

âˆ AOD = âˆ BOC (Vertically opposite angles)

âˆ OAD = âˆ OBC = 90Â° (âˆµ DA and CB are perpendiculars)

DA = CB (Given)

So, by the AAS congruence rule, we obtain:

Î”DAO â‰… Î”CBO

â‡’ AO = BO (By CPCT)

Now, AO + BO = AB

Thus, line segment CD bisects line segment AB at point O.

**Example 3: If two altitudes of a triangle are equal then show that the triangle is isosceles.****Solution:**

Let us consider Î”PQR such that RT and QS are the equal altitudes drawn to the sides PQ and PR respectively.

In Î”PTR and Î”PSQ, we have

âˆ PTR = âˆ PSQ = 90Â°

âˆ TPR = âˆ SPQ (Common angle)

RT = QS (Given)

So, by the AAS congruence rule, we obtain:

Î”PTR â‰… Î”PSQ

âˆ´ PR = PQ (By CPCT)

Thus, Î”PQR is an isosceles triangle.

**Example 1: In the given figure, âˆ DBC = âˆ EAC, âˆ DCA = âˆ ECB and BD = AE. Prove that BC = AC. **

**Solution:**

It is given that âˆ DCA = âˆ ECB.

On adding âˆ ECD to both sides of the above equation, we get:

âˆ DCA + âˆ ECD = âˆ ECB + âˆ ECD

â‡’ âˆ ECA = âˆ DCBâ€¦ (1)

In Î”BDC and Î”AEC, we have:

âˆ DCB = âˆ ECA (From equation 1)

âˆ DBC = âˆ EAC (Given)

BD = AE (Given)

So, by the AAS congruence rule, we have:

Î”BDC â‰… Î”AEC

â‡’ BC = AC (By CPCT)

**Example 2: In the figure, âˆ BAD = âˆ BCE and AB = CB. Prove that Î”ABD â‰…Î”CBE. **

**Solution:**

In Î”AOE and Î”COD, we have:

âˆ EAO = âˆ DCO (Given)

âˆ AOE = âˆ COD (Vertically opposite angles)

âˆ´ âˆ AEO = âˆ CDO â€¦ (1) [By the angle sum property]

Now,

âˆ AEO + âˆ OEB = 180Â° (Linear pair)

âˆ CDO + âˆ ODB = 180Â° (Linear pair)

âˆ´ âˆ AEO + âˆ OEB = âˆ CDO + âˆ ODB

â‡’ âˆ AEO + âˆ OEB = âˆ AEO + âˆ ODB (Using equation (1))

â‡’ âˆ OEB = âˆ ODB

â‡’ âˆ CEB = âˆ ADB â€¦ (2)

In Î”ABD and Î”CBE, we have:

âˆ BAD = âˆ BCE (Given)

âˆ ADB = âˆ CEB (From equation 2)

AB = CB (Given)

So, by the AAS congruence rule, we have:

Î”ABD â‰… Î”CBE

**Example 3: Prove that if two triangles are congruent then their corresponding altitudes are equal.****Solution:**

Let PS and XT be corresponding altitudes of congruent triangles Î”PQR and Î”XYZ.

We have

Î”PQR â‰… Î”XYZ

âˆ´ PQ = XY …(1) (By CPCT)

âˆ PQR = âˆ XYZ …(2) (By CPCT)

In Î”PQS and Î”XYT, we have

âˆ PSQ = âˆ XTY = 90Â°

âˆ PQR = âˆ XYZ [From (2)]

PQ = XY [From (1)]

So, by the AAS congruence rule, we have:

Î”PQS â‰… Î”XYT

âˆ´ PS = XT (By CPCT)

Thus, the corresponding altitudes of congruent triangles are equal.

**Example 1: Are the following triangles congruent? **

**Solution:**

In Î”ABC and Î”PQR, we have:

AB = PQ = 4.6 cm

BC = QR = 5.3 cm

But AC â‰ PR

Therefore, Î”ABC and Î”PQR are not congruent.

**Example 2: ABCD is a rectangle with AC as one of its diagonals. Prove that the triangles formed on the two sides of diagonal AC are congruent.****Solution:**

The required rectangle ABCD with AC as its diagonal can be drawn as is shown.

In Î”ABC and Î”CDA, we have:

AB = CD (âˆµ Opposite sides of a rectangle are equal)

BC = DA (âˆµ Opposite sides of a rectangle are equal)

CA = AC (Common side)

Therefore, by the SSS congruence rule, we have:

Î”ABC â‰… Î”CDA

Thus, the triangles formed on the two sides of diagonal AC are congruent.

**Example 1: The given Î”ABC is isosceles with AB = AC. AD is a median of the triangle. Prove that AD is perpendicular to BC.**

**Solution:**

In Î”ABD and Î”ACD, we have:

AB = AC (Given)

BD = DC (âˆµ D is the midpoint of BC)

AD = AD (Common side)

Therefore, by the SSS congruence rule, we obtain:

Î”ABD â‰… Î”ACD

â‡’ âˆ ADB = âˆ ADC (By CPCT)

Also, âˆ ADB and âˆ ADC form a linear pair.

So, âˆ ADB + âˆ ADC = 180Â°

â‡’ âˆ ADB + âˆ ADB = 180Â° (âˆµ âˆ ADB = âˆ ADC)

â‡’ 2âˆ ADB = 180Â°

â‡’ âˆ ADB = 90Â°

Thus, âˆ ADB = âˆ ADC = 90Â°, which means that AD is perpendicular to BC.

**Example 2: ABCD is a parallelogram. If the diagonals of ABCD are equal, then find the measure of âˆ ABC.****Solution:**

The given parallelogram ABCD with equal diagonals AC and BD is shown below.

In parallelogram ABCD, we have:

AB = CD and AD = BC (âˆµ Opposite sides of a parallelogram are equal)

In Î”ADB and Î”BCA, we have:

AD = BC (Proved above)

BD = AC (Given)

BA = AB (Common side)

So, by the SSS congruence rule, we have:

Î”ADB â‰… Î”BCA

â‡’ âˆ BAD = âˆ ABC â€¦ (1) [By CPCT]

Now, AD is parallel to BC and the transversal AB intersects them at A and B respectively.

We know that the sum of the interior angles on the same side of a transversal is

supplementary.

âˆ´ âˆ BAD + âˆ ABC = 180Â°

â‡’ âˆ ABC + âˆ ABC = 180Â° (By equation 1)

â‡’ 2âˆ ABC = 180Â°

â‡’ âˆ ABC = 90Â°

**Example 1: In the given figure, Î”ABC and Î”DBC are isosceles with AB = AC and DB = DC.****Prove that AD is the perpendicular bisector of BC. **

**Solution:**

In Î”ABD and Î”ACD, we have:

AB = AC (Given)

DB = DC (Given)

AD = AD (Common side)

So, by the SSS congruence rule, we have:

Î”ABD â‰… Î”ACD

â‡’ âˆ BAE = âˆ CAE â€¦ (1) [By CPCT]

In Î”BAE and Î”CAE, we have:

AB = AC (Given)

âˆ BAE = âˆ CAE (From equation 1)

AE = AE (Common side)

So, by the SAS congruence rule, we have:

Î”BAE â‰… Î”CAE

â‡’ BE = CE and âˆ BEA = âˆ CEA (By CPCT)

We know that âˆ BEA + âˆ CEA = 180Â° as they form a linear pair.

So, 2âˆ BEA = 180Â° (Proved above that âˆ BEA = âˆ CEA)

â‡’ âˆ BEA = 90Â°

Therefore, âˆ BEA = âˆ CEA = 90Â°

Since BE = CE and âˆ BEA = âˆ CEA = 90Â°, AD is the perpendicular bisector of BC.

**Example 2: O is a point inside a square ABCD such that it is at an equal distance from points B and D. Prove that points A, O and C are collinear.****Solution:**

The square with the given specifications is drawn as is shown.

Construction: Join point O to the vertices of the square.

In Î”AOD and Î”AOB, we have:

AD = AB (Sides of a square)

AO = AO (Common side)

OD = OB (Given)

So, by the SSS congruence rule, we have:

Î”AOD â‰… Î”AOB

â‡’ âˆ 1 = âˆ 2â€¦ (1) [By CPCT]

Similarly, Î”DOC â‰… Î”BOC

â‡’ âˆ 3 = âˆ 4 â€¦ (2) [By CPCT]

We know that:

âˆ 1 + âˆ 2 + âˆ 3 + âˆ 4 = 360Â°

â‡’ 2âˆ 2 + 2âˆ 3 = 360Â° (From equations 1 and 2)

â‡’ âˆ 2 + âˆ 3 = 180Â°

Thus, âˆ 2 and âˆ 3 form a linear pair. Therefore, AOC is a line; in other words, points A, O and C are collinear.

**Example 1: Î”ABC and Î”LMN are right-angled at âˆ ABC and âˆ LMN respectively. In Î”ABC, AB = 2.5 cm and AC = 4.5 cm. In Î”LMN, LN = 5.5 cm and LM = 2.5 cm. Examine whether the two triangles are congruent.****Solution: **On the basis of the given information, the two triangles can be drawn as is shown.

In Î”ABC and Î”LMN, we have:

âˆ ABC = âˆ LMN (Right angles)

AB = LM = 2.5 cm (Given)

But AC â‰ LN

Hence, Î”ABC and Î”LMN are not congruent.

**Example 2: Find the value of x if the shown triangles ABC and DEF are congruent. **

**Solution:**

It is given that Î”ABC â‰… Î”DEF.

When two triangles are congruent, their corresponding sides are equal.

âˆ´ AC = DF = 5.9 cm

Thus, the value of x is 5.9 cm.

**Example 1: In the given Î”ABC, D is the midpoint of side BC. The perpendiculars DX and DY drawn from point D to sides AB and BC respectively are of the same length. Prove that DX ****and DY make the same angle with BC. IMG****Solution: **On comparing Î”DXB and Î”DYC, we get:

DX = DY (Given)

âˆ DXB = âˆ DYC = 90Â° (âˆµ DX and DY are perpendiculars)

BD = CD (âˆµ D is the midpoint of BC)

Thus, by the RHS congruence rule, we have:

Î”DXB â‰… Î”DYC

â‡’ âˆ BDX = âˆ CDY (By CPCT)

Thus, the perpendiculars DX and DY make the same angle with side BC.

**Example 2: In the given Î”ABC, AD, BE and CF are the altitudes. If the three altitudes are equal, then prove that the triangle is equilateral. **

**Solution:**

In Î”BEC and Î”CFB, we have:

BC = CB (Common side)

BE = CF (Given)

âˆ BEC = âˆ CFB = 90Â° (âˆµ BE and CF are altitudes)

So, by the RHS congruence rule, we obtain:

Î”BEC â‰… Î”CFB

â‡’ âˆ BCE = âˆ CBF (By CPCT)

â‡’ âˆ CBA = âˆ BCA

â‡’ AC = AB â€¦ (1) [âˆµ Sides opposite equal angles are equal]

Similarly, we can prove that Î”ADB â‰… Î”BEA.

â‡’ âˆ DBA = âˆ BAE (By CPCT)

â‡’ âˆ CBA = âˆ BAC

â‡’ AC = BC â€¦ (2) [âˆµ Sides opposite equal angles are equal]

From equations (1) and (2), we get:

AB = BC = AC

Hence, Î”ABC is equilateral.

**Example 1: If Z is a point equidistant from two lines m and n intersecting at point A, then prove that AZ bisects the angle between m and n.****Solution:**

The following figure can be drawn as per the given information.

**Construction:** ZB and ZC are perpendiculars drawn from point Z to

lines m and n respectively.

Since Z is equidistant from m and n, we have:

ZB = ZC

In Î”ZBA and Î”ZCA, we have:

ZB = ZC (Proved above)

âˆ ZBA = âˆ ZCA = 90Â° (âˆµ ZB and ZC are perpendiculars)

ZA = ZA (Common side)

So, by the RHS congruence rule, we have:

Î”ZBA â‰… Î”ZCA

â‡’ âˆ ZAB = âˆ ZAC (By CPCT)

Now, âˆ ZAB + âˆ ZAC = âˆ BAC

Therefore, AZ bisects the angle between lines m and n.

**Example 2: In a Î”ABC, BD = DC. If the perpendiculars from point D to sides AB and AC are equal, then prove that AB = AC.****Solution: **The triangle with the given specifications is drawn below.

In Î”ABC, D is the midpoint of BC. Also, DE and DF are the perpendiculars from D to AB and

AC respectively.

In Î”DEB and Î”DFC, we have:

âˆ DEB = âˆ DFC = 90Â° (âˆµ DE and DF are perpendiculars)

DB = DC (Given)

DE = DF (Given)

So, by the RHS congruence rule, we obtain:

Î”DEB â‰… Î”DFC

â‡’ âˆ DBE = âˆ DCF (By CPCT)

â‡’ âˆ CBA = âˆ BCA

â‡’ AC = AB (As sides opposite to equal angles are equal)

**Example 1: In the given Î”ABC, AB = AC. What is the measure of âˆ BAC? **

**Solution:**

Î”ABC is an isosceles triangle with AB = AC.

By the property of isosceles triangles, we obtain:

âˆ ABC = âˆ ACB

â‡’ âˆ ACB = 50Â°

By the angle sum property, we have:

âˆ ABC + âˆ ACB + âˆ BAC = 180Â°

â‡’ 50Â° + 50Â° + âˆ BAC = 180Â°

â‡’ âˆ BAC = 180Â° âˆ’ 100Â°

â‡’ âˆ BAC = 80Â°

**Example 2: Ram lives in a triangular tree house. The measure of each base angle is the same and the length of the base is 4 m. He wants to cover the slant sides with metal sheets. If the boundary of the triangular structure is 10 m, then what length of sheet does Ram need to cover the slant sides (irrespective of the width of the sheet).****Solution:**

It is given that the base angles are equal; so, the triangular structure is isosceles. Hence, the

slant sides are equal.

Now, we know that:

Base = 4 m

Boundary of the triangular structure = Perimeter of the isosceles triangle = 10 m

â‡’ 10 m = 4 m + 2 Ã— (Measure of each slant side)

â‡’ 2 Ã— (Measure of each slant side) = 10 m âˆ’ 4 m

â‡’ 2 Ã— (Measure of each slant side) = 6 m

Hence, Ram needs 6 m of metal sheet to cover the slant sides of the triangular structure.

**Example 1: Find the missing angles in the following triangles.**

**Solution:**

1. In Î”ABC, AB = AC = 2.5 cm

Since the angles opposite equal sides of a triangle are equal, we obtain:

âˆ ABC = âˆ ACB

Let âˆ ABC = âˆ ACB = x

By the angle sum property of triangles, we have:

âˆ ABC + âˆ ACB + âˆ BAC = 180Â°

â‡’ x + x + 80Â° = 180Â°

â‡’ 2x = 100Â°

â‡’ x = 50Â°

Thus, âˆ ABC = âˆ ACB = 50Â°

2. In Î”PQR, PQ = QR = 5 cm and PR = 3 cm

Since PQ = QR, we obtain

âˆ QPR = âˆ PRQ = 70Â°

**Example 2: The shown Î”ABC is isosceles with AB = AC. Find the measures of âˆ BAC, âˆ ABC and âˆ ACB. **

**Solution:Â **âˆ ACB and exterior angle âˆ ACD form a linear pair.

âˆ´ âˆ ACB + âˆ ACD = 180Â°

â‡’ x + 2x = 180Â°

â‡’ 3x = 180Â°

â‡’ âˆ´ x = 60Â°

So, âˆ ACB = 60Â° and âˆ ACD = 120Â°

Î”ABC is isosceles.

âˆ´ âˆ ABC = âˆ ACB = 60Â° (âˆµ Angles opposite equal sides are equal)

Now, by the exterior angle property, we have:

âˆ ACD = âˆ BAC + âˆ ABC

â‡’ 120Â° = âˆ BAC + 60Â°

â‡’ âˆ BAC = 60Â°

Thus, âˆ BAC = âˆ ABC = âˆ ACB = 60Â°.

**Example 3: Prove that in an isosceles triangle, the angle bisector of the apex angle is theÂ perpendicular bisector of the base.****Solution:****Given:** ABC is a triangle in which AB = AC and apex angle âˆ A.**To Prove:** AD is perpendicular bisector of BC and BD = DC.**Construction:** Draw an angle bisector AD from A on BC.**Proof:Â **

In Î”ABD and Î”ACD,

AB = AC (Given)

AD = AD (Common side)

âˆ BAD = âˆ CAD (AD is a bisector of âˆ A)

So, by SAS congruence criterion,

Î” ADBÂ â‰…Â Î” ADC

â‡’BD = DC and âˆ ADB =âˆ ADC (CPCT)

Now,

âˆ ADB + âˆ ADC= 180Â° (Linear Pair)

â‡’âˆ ADB + âˆ ADB = 180Â°

â‡’ 2âˆ ADB = 180Â°

â‡’Â âˆ ABD =Â 180Â°/2 = 90Â°

Thus, AD is the perpendicular bisector of BC.

**Example 1: Î”ABD is isosceles (figure not drawn to scale) with AB = BD, while Î”BCD is isosceles with BC = BD. Also, âˆ BDC measures 70Â° and âˆ ABC measures 90Â°. What is theÂ measure of âˆ ADC? **

**Solution:**

In Î”BCD, BC = BD.

âˆ´ âˆ BCD = âˆ BDC = 70Âº (âˆµ Angles opposite equal sides are equal)

In Î”BCD, by the angle sum property, we obtain:

âˆ BCD + âˆ BDC + âˆ CBD = 180Âº

â‡’ 70Âº + 70Âº + âˆ CBD = 180Âº

â‡’ 140Âº + âˆ CBD = 180Âº

â‡’ âˆ CBD = 40Âº

It is given that âˆ ABC = 90Âº.

â‡’ âˆ ABD + âˆ CBD = 90Âº

â‡’ âˆ ABD + 40Âº = 90Âº

â‡’ âˆ ABD = 50Âº

In Î”ABD, we have:

AB = BD

âˆ´ âˆ BAD = âˆ BDA = x (âˆµ Angles opposite equal sides are equal)

In Î”ABD, by the angle sum property, we obtain:

âˆ ABD + âˆ BAD + âˆ BDA = 180Âº

â‡’ 50Âº + x + x = 180Âº

â‡’ 2x = 130Âº

â‡’ x = 65Âº

So, âˆ BAD = âˆ BDA = 65Âº

Now,

âˆ ADC = âˆ BDA + âˆ CDB

â‡’ âˆ ADC = 65Âº + 70Âº

â‡’ âˆ ADC = 135Âº

**Example 2: In the given figure, Î”PQR is isosceles with PQ = PR. Side SU is extended up to point V such that ST = SV. If âˆ QTS = âˆ RUS = 90Â°, prove that QS bisects âˆ TSU and henceÂ show that Î”QTS â‰…Î”QVS. ****Solution:**

In Î”PQR, we have:

PQ = PR (Given)

âˆ´ âˆ PQR = âˆ PRQ (âˆµ Angles opposite equal sides are equal)

It is given that âˆ QTS = 90Â°.

Using the angle sum property in Î”QTS, we obtain:

âˆ TQS + âˆ TSQ + âˆ QTS = 180Â°

â‡’ âˆ TQS + âˆ TSQ + 90Â° = 180Â°

â‡’ âˆ TSQ = 90Â° âˆ TQS â€¦ (1)

It is given that âˆ RUS = 90Â°.

Using the angle sum property in Î”RUS, we obtain:

âˆ SRU + âˆ RUS + âˆ RSU = 180Â°

â‡’ âˆ SRU + 90Â° + âˆ RSU = 180Â°

â‡’ âˆ RSU = 90Â° –Â âˆ SRU â€¦ (2)

âˆ SRU = âˆ PRQ (Vertically opposite angles)

Also, âˆ PQR = âˆ PRQ

âˆ´ âˆ SRU = âˆ PQR â€¦ (3)

Now, from equations (2) and (3), we get:

âˆ RSU = 90Â° –Â âˆ PQR

â‡’ âˆ RSU = 90Â° –Â âˆ TQS â€¦ (4) [âˆµ âˆ PQR = âˆ TQS]

From equations (1) and (4), we get:

âˆ TSQ = âˆ RSU

Hence, QS bisects âˆ TSU.

Now, consider Î”QTS and Î”QVS.

QS = QS (Common side)

âˆ TSQ = âˆ QSV (âˆµ âˆ TSQ = âˆ RSU and âˆ RSU = âˆ QSV)

ST = SV (Given)

Thus, by the SAS congruency rule, we get:

Î”QTS â‰… Î”QVS

**Example 1: In a Î”ABC, âˆ BAC = 2x and âˆ ABC = âˆ ACB = x. Find the value of x and hence ****show that AB = AC.****Solution:**

It is given that âˆ BAC = 2x and âˆ ABC = âˆ ACB = x.

By applying the angle sum property, we obtain:

âˆ BAC + âˆ ABC + âˆ ACB = 180Â°

â‡’ 2x + x + x = 180Â°

â‡’ 4x = 180Â°

â‡’ x = 45Â°

So, âˆ BAC = 2x = 2 Ã— 45Â° = 90Â° and âˆ ABC = âˆ ACB = x = 45Â°

The given triangle can be drawn as is shown. IMG

We know that the sides opposite equal angles of a triangle are equal.

âˆ´ AB = AC

Hence, Î”ABC is isosceles with AB = AC.

**Example 2: In the given Î”ABC, âˆ ABC = âˆ ACB and the perimeter is 11 cm. Find the length of the base of the triangle. ****Solution:** It is given that âˆ ABC = âˆ ACB and AB = 3 cm.

We know that the sides opposite equal angles of a triangle are equal.

âˆ´ AC = AB = 3 cm

Perimeter of Î”ABC = AB + BC + AC

â‡’ 11 cm = 3 cm + BC + 3 cm

â‡’ BC = 5 cm

Thus, the length of the base of Î”ABC is 5 cm.

**Example 1: In the given Î”ABC, âˆ ABD = âˆ ACD and AD is perpendicular to BC. Prove that AD bisects âˆ BAC. ****Solution:**

In Î”ABD and Î”ACD, we have:

âˆ ABD = âˆ ACD (Given)

âˆ ADB = âˆ ADC = 90Â°

AB = AC (âˆµ Sides opposite equal angles are equal)

Thus, by the AAS congruence rule, we have:

Î”ABD â‰… Î”ACD

âˆ´ âˆ BAD = âˆ CAD (By CPCT)

Thus, AD bisects âˆ BAC.

**Example 2: In the given Î”ABC, âˆ ABC = âˆ ACB. Side BA is produced up to point D such that AB = AD. Prove that âˆ BCD is a right angle. ****Solution:**

In Î”ABC, we have:

âˆ ABC = âˆ ACB â€¦ (1) [Given]

âˆ´ AB = AC (âˆµ Sides opposite equal angles are equal)

Now,

AB = AD (Given)

AD = AC (âˆµ AB = AC)

Thus, in Î”ADC, we have:

âˆ ACD = âˆ ADC â€¦ (2) (âˆµ Angles opposite equal sides are equal)

On adding equations (1) and (2), we get:

âˆ ACB + âˆ ACD = âˆ ABC + âˆ ADC

â‡’ âˆ BCD = âˆ ABC + âˆ BDC (âˆµ âˆ ADC = âˆ BDC)

On adding âˆ BCD to both sides of the equation, we get:

âˆ BCD + âˆ BCD = âˆ ABC + âˆ BDC + âˆ BCD

â‡’ 2âˆ BCD = 180Â° (By the angle sum property)

â‡’ âˆ BCD = 90Â°

Hence, âˆ BCD is a right angle.

**Example 1: The shown Î”PQR is isosceles with PQ = PR. QL and RM are the respective medians from vertices Q and R to sides PR and PQ. Prove that the medians have the same ****length.****Solution:**

It is given that PQ = PR.

Also, QL and RM are medians.

âˆ´ PL = LR and PM = MQ

So,

PL + LR = PR

LR + LR = PR

2LR = PR

LR = PR/2 â€¦ (1)

Similarly, MQ = PQ/2 â€¦ (2)

Since PQ = PR, using equations (1) and (2), we obtain:

LR = MQ â€¦ (3)

In Î”QRL and Î”RQM, we have:

LR = MQ (From (3))

âˆ LRQ = âˆ MQR (âˆµ Angles opposite equal sides are equal)

QR = RQ (Common side)

âˆ´ Î”QRL â‰… Î”RQM (By the SAS congruence criterion)

â‡’ QL = RM (By CPCT)

Thus, the medians QL and RM have the same length.

**Example 2: In a right-angled triangle ABC, âˆ ACB = 2âˆ BAC. Prove that AC = 2BC.****Solution: **The given Î”ABC can be drawn as is shown.

Construction: Produce CB up to point D such that BD = BC. Join point A to point D. IMG

In Î”ABD and Î”ABC, we have:

BD = BC (By construction)

âˆ ABD = âˆ ABC = 90Â°

AB = AB (Common side)

So, by the SAS congruence rule, we obtain:

Î”ABD â‰… Î”ABC

â‡’ AD = AC and âˆ DAB = âˆ BAC (By CPCT)

Let âˆ BAC be x. Then, âˆ DAB will also be x.

Now, âˆ DAC = âˆ DAB + âˆ BAC

â‡’ âˆ DAC = x + x

â‡’ âˆ DAC = 2x

â‡’ âˆ DAC = âˆ ACB (âˆµ âˆ ACB = 2âˆ BAC = 2x)

â‡’ DC = AD (âˆµ Sides opposite equal angles are equal)

Since BC = DB, we have:

DC = 2BC

â‡’ 2BC = AD

â‡’ 2BC = AC (âˆµ We have proved AD = AC)

**Example 1: In the given Î”PQR, PQ is greater than PR. Also, QS and RS are the respective bisectors of âˆ PQR and âˆ PRQ. Prove that âˆ SRQ > âˆ SQR.****Solution:**

In Î”PQR, we have:

PQ > PR (Given)

â‡’ âˆ PRQ > âˆ PQR (âˆµ Angle opposite longer side is greater)

â‡’Â âˆ PRQ/2Â >Â âˆ PRQ/2

â‡’ âˆ SRQ > âˆ SQR (âˆµ RS bisects âˆ PRQ and QS bisects âˆ PQR)

**Example 2: ABC is an isosceles triangle with AB = AC and AB < BC. Prove that âˆ BAC > âˆ ABC.****Solution:**

Consider the following Î”ABC in which AB = AC and AB < BC.

In Î”ABC, we have:

âˆ ABC = âˆ ACB â€¦ (1) [âˆµ Angles opposite equal sides AB and AC are equal]

By the triangle inequality theorem, we have:

âˆ ACB < âˆ BAC â€¦ (2) [âˆµ Angle opposite longer side BC is greater]

Thus, from (1) and (2), we obtain:

âˆ BAC > âˆ ABC

**Example 1: In the shown Î”ABC, AB = 4 cm, BC = 5 cm and the perimeter is 16 cm.****Determine the smallest and greatest angles of the triangle. ****Solution:**

In Î”ABC, we have:

AB = 4 cm and BC = 5 cm (Given)

Perimeter = 16 cm (Also given)

â‡’ AB + BC + CA = 16 cm (âˆµ Perimeter is the sum of all sides)

â‡’ 4 cm + 5 cm + CA = 16 cm

â‡’ CA = 7 cm

Now, CA = 7 cm is the longest side of Î”ABC. Thus, the angle opposite it, i.e., âˆ ABC is the

greatest angle of the triangle.

âˆ´ âˆ ABC > âˆ BAC and âˆ ABC > âˆ ACB â€¦ (1)

Also, BC > AB

âˆ´ âˆ BAC > âˆ ACB â€¦ (2)

Thus, from (1) and (2), we can conclude that âˆ ABC is the greatest angle and âˆ ACB is the smallest angle in Î”ABC.

**Example 2: Suppose AB is the longest side and CD the shortest side of the given quadrilateral ABCD. Then, prove that âˆ BCD > âˆ BAD. ****Solution:**

In Î”ABC, AB is the longest side.

So, AB > BC

â‡’ âˆ 1 > âˆ 2 â€¦ (1) [âˆµ Angle opposite longer side is greater]

In Î”ADC, CD is the shortest side.

So, AD > CD

â‡’ âˆ 4 > âˆ 3 â€¦ (2) [âˆµ Angle opposite longer side is greater]

On adding (1) and (2), we get:

âˆ 1 + âˆ 4 > âˆ 2 + âˆ 3

â‡’ âˆ BCD > âˆ BAD

**Example 1: In the given figure, PQ = PR. Show that âˆ PRS > âˆ PSR. ****Solution:**

In Î”PQR, we have:

PQ = PR

â‡’ âˆ PRQ = âˆ PQR â€¦ (1) [âˆµ Angles opposite equal sides are equal]

In Î”PSQ, SQ is produced to R.

So, âˆ PQR = âˆ PSQ + âˆ SPQ (By the exterior angle property)

â‡’ âˆ PQR > âˆ PSQ â€¦ (2)

From (1) and (2), we obtain:

âˆ PRQ >âˆ PSQ

â‡’ âˆ PRS > âˆ PSR (âˆµ âˆ PRQ = âˆ PRS and âˆ PSQ = âˆ PSR)

**Example 2: In a Î”ABC, AB > AC and AD is the bisector of âˆ BAC. Show that âˆ ADB > âˆ ADC.****Solution:Â **Let the following Î”ABC be the given triangle such that AB > AC. Also, AD is the bisector of âˆ BAC.

In Î”ABC, we have:

AB > AC (Given)

â‡’ âˆ ACB > âˆ ABC (âˆµ Angle opposite longer side is greater)

On adding âˆ 1 to both sides, we get:

âˆ ACB + âˆ 1 > âˆ ABC + âˆ 1

â‡’ âˆ ACB + âˆ 2 > âˆ ABC + âˆ 1 â€¦ (1) [âˆµ AD bisects âˆ BAC; âˆ 1 = âˆ 2]

By the exterior angle property, we have:

âˆ ADB = âˆ ACB + âˆ 2 â€¦ (2)

Similarly, âˆ ADC = âˆ ABC + âˆ 1 â€¦ (3)

Thus, by using (1), (2) and (3), we can conclude that:

âˆ ADB > âˆ ADC

**Example 1: The angles of a triangle are in the ratio 1 : 2 : 3. Find the greatest angle and identify the longest side of the triangle.****Solution:**

It is given that the angles of the triangle, say Î”ABC, are in ratio 1 : 2 : 3.

Let the angles be x, 2x and 3x, as is shown in the figure.

By the angle sum property of triangles, we have:

âˆ BAC + âˆ ABC + âˆ ACB = 180Â°

â‡’ x + 3x + 2x = 180Â°

â‡’ 6x = 180Â°

â‡’ x = 30Â°

So, 2x = 2 Ã— 30Â° = 60Â° and 3x = 3 Ã— 30Â° = 90Â°

Thus, the greatest angle in the given triangle is âˆ ABC, i.e., 90Â°.

Now, we know that the side opposite the greater angle is longer. In Î”ABC, side AC is opposite the greatest angle; hence, it is the longest.

**Example 2: Which side of the given triangle is the shortest? ****Solution:**

In order to find the shortest side of Î”DEF, we need to figure out the smallest angle of the

triangle. This is because the smallest side is opposite the smallest angle.

By the angle sum property of triangles, we have:

âˆ EDF + âˆ DEF + âˆ EFD = 180Â°

â‡’ 48Â° + 87Â° + âˆ EFD = 180Â°

â‡’ 135Â° + âˆ EFD = 180Â°

â‡’ âˆ EFD = 45Â°

Clearly, âˆ EFD has the smallest measure in the given triangle. So, the side opposite it, i.e., DE is the shortest side of the triangle.

**Example 1: In the given Î”ABC, âˆ ABD > âˆ ACD and AD is the bisector of âˆ BAC.****Prove that:**

1. AB < AC

2. âˆ ADB < âˆ ADC IMG**Solution:**

1. It is given that âˆ ABD > âˆ ACD.

âˆ´ AC > AB (âˆµ Side opposite greater angle is longer)

Or, AB < AC

2. In Î”ABD, we have:

âˆ ABD + âˆ BAD + âˆ ADB = 180Â° (By the angle sum property)

â‡’ âˆ ACD + âˆ BAD + âˆ ADB < 180Â° (âˆµ âˆ ABD > âˆ ACD)

â‡’ âˆ ACD + âˆ CAD + âˆ ADB < 180Â° (âˆµ AD bisects âˆ BAC; âˆ BAD = âˆ CAD)

â‡’ âˆ ADB < 180Â° âˆ’ (âˆ ACD +âˆ CAD)

â‡’ âˆ ADB < âˆ ADC (By the angle sum property in Î”ADC)

**Example 2: Prove that of all the line segments that can be drawn to a given line from a point lying outside the line, the perpendicular line segment is the shortest.****Solution:**

Let there be a straight line m, a point P lying outside the line and a point M lying on the line.

Also, PM âŠ¥ m and N is any point other than M on m.

Since PM is perpendicular to m, âˆ PMN = 90Â°.

In Î”PMN, we have:

â‡’ âˆ PMN + âˆ MPN + âˆ PNM = 180Â° (By the angle sum property)

â‡’ 90Â° + âˆ MPN + âˆ PNM = 180Â°

â‡’ âˆ MPN + âˆ PNM = 90Â°

â‡’ âˆ PNM < 90Â°

â‡’ âˆ PNM < âˆ PMN

â‡’ PM < PN (âˆµ Side opposite greater angle is longer)

Thus, PM is the shortest of all line segments from point P to line m.

**Example 1:** In the given Î”PQR, âˆ PQR = 50Â°, âˆ PRQ = 20Â° and âˆ PSQ = 30Â°. Prove that QS > SR. **Solution:**

In Î”PQS, we have:

âˆ PQR = 50Â° and âˆ PSQ = 30Â° (Given)

By using the angle sum property in Î”PQS, we obtain:

âˆ PQR + âˆ PSQ + âˆ QPS = 180Â°

â‡’ 50Â° + 30Â° + âˆ QPS = 180Â°

â‡’ âˆ QPS = 180Â° âˆ’ (50Â° + 30Â°)

â‡’ âˆ QPS = 100Â°

So, âˆ QPS > âˆ PQR > âˆ PSQ

Since the side opposite the greater angle is longer, we get:

QS > PS > PQ â€¦ (1)

âˆ PSQ and âˆ PSR form a linear pair.

So, âˆ PSQ + âˆ PSR = 180Â°

â‡’ 30Â° + âˆ PSR = 180Â°

â‡’ âˆ PSR = 180Â° âˆ’ 30Â°

â‡’ âˆ PSR = 150Â°

In Î”PSR, we have:

âˆ PSR + âˆ SPR + âˆ PRQ = 180Â° (By the angle sum property)

â‡’ 150Â° + âˆ SPR + 20Â° = 180Â°

â‡’ âˆ SPR = 180Â° âˆ’ (150Â° + 20Â°)

â‡’ âˆ SPR = 10Â°

So, âˆ PSR > âˆ PRQ > âˆ SPR

Since the side opposite the greater angle is longer, we get:

PR > PS > SR â€¦ (2)

By using (1) and (2), we can conclude that:

QS > SR

**Example 2: Prove that the difference between any two sides of a triangle is less than the third side.****Solution:**

We have to prove that the difference between any two sides of a triangle (let us say Î”ABC) is less than the third side of the triangle. We will do so by showing that in Î”ABC, BC âˆ’ AC <

AB.**Construction:** Extend side BA up to point D such that AD = AC. Join C to D. IMG

In Î”ACD, we have:

AD = AC â€¦ (1)

We know that in an isosceles triangle, the angles opposite equal sides are equal.

âˆ´ âˆ ACD = âˆ ADC

â‡’ âˆ ACD + âˆ ACB > âˆ ADC

â‡’ âˆ BCD > âˆ ADC

We know that the side opposite the greater angle is longer. So, we obtain, BD > BC

â‡’ AB + AD > BC (âˆµ BD = AB + AD)

â‡’ AB + AC > BC (Using equation 1)

â‡’ BC âˆ’ AC < AB

Similarly, we can prove that AB âˆ’ BC < AC and AC âˆ’ AB < BC.

Thus, we have proved that the difference between any two sides of a triangle is less than the third side.