# Frank Solutions for Chapter 12 Isosceles Triangles Class 9 Mathematics ICSE

### Exercise 12.1

1. Find the angles of an isosceles triangle whose equal angles and the non-equal angles are in the ratio 3:4.

Given that,

The equal angles and the non-equal angles are in the ratio 3:4

Let the equal angles be 3x each,

So, non- equal angle is 4x

We know that,

Sum of angles of a triangle = 180°

Hence,

3x + 3x + 4x = 180°

⇒ 10x = 180°

We get,

x = 18°

Therefore,

3x = 3 × 18° = 54° and

4x = 4 × 18° = 72°

Hence,

The angles of a triangle are 54°, 54° and 72°

2. Find the angles of an isosceles triangle which are in the ratio 2:2:5

The equal angles and the non-equal angle are in the ratio 2:2:5

Let equal angles be 2x each

So, non-equal angle is 5x

We know that,

Sum of angles of a triangle = 180°

2x + 2x + 5x = 180°

9x = 180°

x = 20°

Therefore,

2x = 2 × 20° = 40°

5x = 5 × 20° = 100°

Hence, the angles of a triangle are 40°, 40° and 100°

3. Each equal angle of an isosceles triangle is less than the third angle by 15°. Find the angles.

Let equal angles of the isosceles triangle be x each

Therefore, non-equal angle is x + 15°

We know that,

Sum of angles of a triangle = 180°

x + x + (x + 15°) = 180°

⇒ 3x + 15° = 180°

⇒ 3x = 180° – 15°

⇒ 3x = 165°

We get,

x = 55°

So,

(x + 15°) = 55° + 15° = 70°

Hence, the angles of a triangle are 55°, 55° and 70°

4. Find the interior angles of the following triangles

(a)

(b)

(c)

(d)

(a)

In △ABC,

∠A = 110°

AB = AC

∠C = ∠B (angles opposite to two equal sides are equal)

Now,

By angle sum property,

∠A + ∠B + ∠C = 180°

⇒ ∠A + ∠B + ∠B = 180°

⇒ 110° + 2∠B = 180°

⇒ 2∠B = 180° – 110°

⇒ 2∠B = 70°

We get,

∠B = 35°

∠C = 35°

Hence,

The interior angles are ∠B = 35° and ∠C = 35°

(b)

In △ABC,

AB = AC

∠ACB = ∠ABC ...(1) [∵ angles opposite to two equal sides are equal]

Now,

∠ACB + ∠ACD = 180°  [linear pair]

∠ACB = 180° – ∠ACD

∠ACB = 180° – 105°

∠ACB = 75°

So,

∠ABC = 75° [from equation (1)]

Now, in △ABC,

By angle sum property,

∠ABC + ∠ACB + ∠BAC = 180°

75° + 75° + ∠BAC = 180°

150° + ∠BAC = 180°

∠BAC = 180° – 150°

We get,

∠BAC = 30°

Hence,

In △ABC,

∠A = 30°

∠B = 75°

∠C = 75°

(c)

In △ABD,

Given that,

∠ABD = ∠BAD …(angles opposite to two equal sides are equal)

Now,

∠ABD = 37° …(given)

Hence,

By exterior angle property,

∠ADC = 37° + 37°

We get,

AC = DC …(given)

∠ADC = ∠DAC …(angles opposite to two equal sides are equal)

∠DAC = 74°

Now,

∠BAC = ∠BAD + ∠DAC

∠BAC = 37° + 74°

We get,

∠BAC = 111°

In △ABC,

∠BAC + ∠ABC + ∠ACB = 180°

111° + 37° + ∠ACB = 180°

∠ACB = 180° – 111° – 37°

We get,

∠ACB = 32°

Therefore,

The interior angles of △ABC are 37°, 111° and 32°

(d)

In △ACD,

AD = CD …(given)

∠ACD = ∠CAD …(angles opposite to two equal sides are equal)

Now,

∠ACD = 50° …(given)

By exterior angle property,

∠ADB = 500 + 500

AD = BD …(given)

∠DBA = ∠DAB …(angles opposite to two equal sides are equal)

Also,

∠ADB + ∠DBA + ∠DAB = 180°

⇒ 100° + 2∠DBA = 180°

⇒ 2∠DBA = 180° – 100°

⇒ 2∠DBA = 80°

We get,

∠DBA = 40°

∠DAB = 40°

∠BAC = ∠DAB + ∠CAD

⇒ ∠BAC = 40° + 50°

⇒ ∠BAC = 90°

Therefore, the interior angles of △ABC are 50°, 90° and 40°

5. Side BA of an isosceles triangle ABC is produced so that AB = AD. If AB and AC are the equal sides of the isosceles triangle, prove that BCD is a right angle.

Let ∠ABC = x

Hence,

∠BCA = x (since AB = AC)

In △ABC,

∠ABC + ∠BCA + ∠BAC = 180° …(1)

But

∠BAC + ∠DAC = 180° …(2)

From equations (1) and (2)

∠ABC + ∠BCA + ∠BAC = ∠BAC + ∠DAC

⇒ ∠DAC = ∠ABC + ∠BCA

⇒ ∠DAC = x + x

We get,

∠DAC = 2x

Let ∠ADC = y,

Hence,

∠DCA = y (since AD = AC)

Now,

∠ADC + ∠DCA + ∠DAC = 180° …(3)

But ∠BAC + ∠DAC = 180° …(4)

From equations (3) and (4), we get,

∠ADC + ∠DCA + ∠DAC = ∠BAC + ∠DAC

⇒ ∠BAC = ∠ADC + ∠DCA

⇒ ∠BAC = y + y

⇒ ∠BAC = 2y

Now, substituting the value of ∠BAC and ∠DAC in equation (2)

2x + 2y = 180°

⇒ x + y = 90°

⇒ ∠BCA + ∠DCA = 90°

Therefore,

∠BCD is a right angle

6. The bisectors of the equal angles of an isosceles triangle PQR meet at O. If PQ = PR, prove that PO bisects P.

Join PO and produce to meet QR at point S

In △PQS and △PRS

PS = PS (common)

PQ = PR (given)

So,

∠Q = ∠R (angles opposite to two equal sides are equal)

Hence,

△PQS ≅ △PRS

Thus,

∠QPS = ∠RPS

Therefore,

PO bisects ∠P

7. Prove that the medians corresponding to equal sides of an isosceles triangle are equal.

Let △ABC be an isosceles triangle with AB = AC

Let D and E be the mid points of AB and AC respectively

Now,

Join BE and CD

Then BE and CD become the medians of this isosceles triangle

In △ABE and △ACD

AB = AC (given)

AD = AE (D and E are mid points of AB and AC)

∠A = ∠A (common angle)

Hence,

△ABE ≅ △ACD (SAS criteria)

Therefore,

The medians BE and CD are equal i.e. BE = CD

8. DPQ is an isosceles triangle with DP = DQ. A straight line CD bisects the exterior QDR. Prove that DC is parallel to PQ

In △QDP,

DP = DQ

Hence,

∠Q = ∠P (angles opposite to two equal sides are equal)

∠QDR = ∠Q + ∠P

⇒ 2∠QDC = ∠Q + ∠P (DC bisects angle QDR)

⇒ 2∠QDC = ∠Q + ∠Q

We get,

2∠QDC = 2∠Q

Hence,

∠QDC = ∠Q

But these angles are alternate angles

Therefore,

DC || PQ

Hence, proved

9. In a quadrilateral PQRS, PQ = PS and RQ = RS. If P = 50° and R = 110°, find PSR.

In △PQS,

PQ = PS

Therefore,

∠PQS = ∠PSQ (angles opposite to two equal sides are equal)

∠P + ∠PQS + ∠PSQ = 180°

⇒ 50° + 2∠PQS = 180°

⇒ 2∠PQS = 180° – 50°

We get,

2∠PQS = 130°

⇒ ∠PQS = 65°

So,

∠PQS = ∠PSQ = 65° …(1)

In △SRQ,

SR = RQ

Hence,

∠RQS = ∠RSQ (angles opposite to two equal sides are equal)

∠R + ∠RQS + ∠RSQ = 180°

⇒ 110° + 2∠RQS = 180°

⇒ 2∠RQS = 180° – 110°

We get,

2∠RQS = 70°

∠RQS = 35°

So,

∠RQS = ∠RSQ = 35° ….(2)

Adding equations (1) and (2), we get,

∠PSQ + ∠RSQ = 65° + 35°

∠PSR = 100°

10. △ABC is an isosceles triangle with AB = AC. Another triangle BDC is drawn with base BC = BD in such a way that BC bisects B. If the measure of BDC is 70°, find the measures of DBC and BAC.

In △BDC,

∠BDC = 70°

BD = BC

Hence,

∠BDC = ∠BCD (angles opposite to two equal sides are equal)

∠BCD = 70°

Now,

∠BCD + ∠BDC + ∠DBC = 180°

⇒ 70° + 70° + ∠DBC = 180°

⇒ ∠DBC = 180° – 140°

We get,

∠DBC = 40°

∠DBC = ∠ABC  (BC is the angle bisector)

Hence,

∠ABC = 40°

In △ABC,

Since AB = AC, ∠ABC = ∠ACB

Hence,

∠ACB = 40°

⇒ ∠ACB + ∠ABC + ∠BAC = 180°

⇒ 40° + 40° + ∠BAC = 180°

⇒ ∠BAC = 180° – 80°

⇒ ∠BAC = 100°

Therefore, the measure of ∠BAC = 100° and ∠DBC = 40°

11. △PQR is isosceles with PQ = PR. T is the mid-point of QR, and TM and TN are perpendiculars on PR and PQ respectively. Prove that,

(i) TM = TN

(ii) PM = PN and

(iii) PT is the bisector of P

(i) In △PQR,

PQ = PR

Hence,

∠R = ∠Q …(1)

Now,

In △QNT and △RMT

∠QNT = ∠RMT = 90°

∠Q = ∠R [from equation (1)]

QT = TR (given)

Hence,

△QNT ≅ △RMT (AAS criteria)

Therefore,

TM = TN

(ii) Since, △QNT ≅ △RMT

NQ = MR …(2)

But,

PQ = PR …(3) [given]

Now, subtracting (2) from (3), we get,

PQ – NQ = PR – MR

PN = PM

(iii) In △PNT and △PMT

TN = TM (proved)

PT = PT (common)

∠PNT = ∠PMT = 90°

Hence,

△PNT ≅ △PMT

So,

∠NPT = ∠MPT

Therefore,

PT is the bisector of ∠P

12. △PQR is isosceles with PQ = QR. QR is extended to S so that PRS becomes isosceles with PR = PS. Show that PSR: QPS = 1:3

In △PQR,

PQ = QR (given)

∠PRQ = ∠QPR  …(1)

In △PRS,

PR = RS (given)

∠PSR = ∠RPS …(2)

Now,

Adding equations (1) and (2), we get,

∠QPR + ∠RPS = ∠PRQ + ∠PSR

∠QPS = ∠PRQ + ∠PSR …(3)

Now,

In △PRS,

∠PRQ = ∠RPS + ∠PSR

∠PRQ = ∠PSR + ∠PSR [from equation(2)]

∠PRQ = 2∠PSR …(4)

Now,

∠QPS = 2∠PSR + ∠PSR [from equation (3) and (4)]

∠QPS = 3∠PSR

∠PSR/∠QPS = 1/3

Therefore,

∠PSR: ∠QPS = 1: 3

Hence, proved

13. In KLM, KT bisects LKM and KT = TM. If LTK is 800, find the value of LMK and KLM.

In △KTM,

KT = TM (given)

Hence,

∠TKM = ∠TMK …(1)

Now,

∠KTL = ∠TKM + ∠TMK

⇒ 80° = ∠TKM + ∠TKM …[from (1)]

⇒ 80° = 2∠TKM

We get,

∠TKM = 40° = ∠TMK = ∠LMK …(2)

But,

∠TKM = ∠TKL (KT is the angle bisector)

Hence,

∠TKL = 40°

In △KTL,

∠TKL + ∠KTL + ∠KLT = 180°

⇒ 40° + 80° + ∠KLT = 180°

⇒ ∠KLT = 180° – 40° – 80°

We get,

∠KLT = 600 = ∠KLM

Therefore,

∠KLM = 60° and ∠LMK = 40°

14. Equal sides QP and RP of an isosceles PQR are produced beyond P to S and T such that PST is an isosceles triangle with PS = PT. Prove that TQ = SR.

In △PTQ and △PSR

PQ = PR (given)

PT = PS (given)

∠TPQ = ∠SPR (vertically opposite angles)

Hence,

△PTQ ≅ △PSR

Therefore,

TQ = SR

Hence, proved

15. Prove that the bisector of the vertex angle of an isosceles triangle bisects the base perpendicularly.

AB = AC (given)

Hence,

Therefore,

BD = DC and ∠BDA = ∠CDA

But,

∠BDA + ∠CDA = 180°

∠BDA = ∠CDA = 90°

Therefore,

AD bisects BC perpendicularly

Hence, proved

16. In the figure △ABC is isosceles with AB = AC. Prove that:

(i) ∠A : ∠B = 1 : 3

(ii) ∠ADE = ∠BCD

In △DEC,

∠DEC = ∠ADE + ∠A = 2a (ext. Angle to △ADE)

DE = DC

⇒ ∠DEC = ∠DEC = 2a …(ii)

In △BDC, let ∠B = b

DC = BC

⇒ ∠BDC = ∠B = b …(iii)

In △ABC,

∠ADB = ∠ADE + ∠EDC + ∠BDC

180° = a + ∠EDC + b (from (i) and (iii))

∠EDC = 180° - a – b …(iv)

Now again in △DEC

180° = ∠EDC + ∠DCE + ∠DEC (from (ii))

⇒ 180° = ∠EDC + 2a + 2a

⇒ ∠EDC = 180° - 4a …(v)

Equating (iv) and (v)

180° - a – b = 180° - 4a

⇒ 3a = b …(vi)

⇒ a/b = 1/3 = ∠A/∠B

Hence, ∠A : ∠B = 1 : 3

∠ADE = a (from (i))

In △BCD,

∠BCE = b (AB = AC ⇒ ∠B = ∠C) …(vii)

⇒ ∠BCD = ∠BCE - ∠DCE

⇒ ∠BCD = b – 2a (from (vii) and (ii))

But b = 3a (from (vi))

Therefore, ∠BCD = 3a – 2a = a

Hence, ∠ADE = ∠BCD

17. In △ABC, D is mid-point of BC, AD is equal to AC. AC is produced to E, such that CE = AC. Prove that :

(i) ∠ADB = ∠DCE

(ii) AB = CE

AD = AC (given)

Therefore, ∠ADC = ∠ACD …(i)

But ∠ADB + ∠ADC = 180° …(ii)

And ∠ACD + ∠DCE = 180° …(iii)

From (i), (ii) and (iii)

(ii) In △ABD and △DCE

BD = CD (D is mid-point of BC)

∠ADB = ∠DCE (proved)

AD = CE (since AC = AC and AC = CE)

Therefore, △ABD ≅ △DCE

Hence, AB = CE

18. In △XYZ, AY and AZ are the bisectors of ∠Y and ∠Z respectively. The perpendicular bisectors of AY and AZ cut YZ at B and C respectively. Prove that line segment YZ is equal to the perimeter of △ABC.

Let M and N be the points where AY and AZ are bisected.

In △ABM and △BMY

MY = MA (BM bisects AY)

BM = BM (common)

∠BMY = ∠BMA

Therefore, △ABM ≅ △BMY

Hence, YB = AB …(i)

In △ACN and △CNZ

NZ = NA (CN bisects AZ)

CN = CN (common)

Therefore, △ACN ≅ △CNZ

Hence, CZ = AC …(ii)

YZ = YB + BC + CZ

Substituting from (i) and (ii)

YZ = YB + BC + AC

Hence, YZ is equal to the perimeter of △ABC

19. △PQR is an isosceles triangle with PQ = PR. QR is extended to S and ST is drawn perpendicular to QP produced, and SN is perpendicular to PR produced. Prove that QS bisects ∠TSN.

In △PQR, let ∠PQR = x

PQ = PR

⇒ ∠PQR = ∠PRQ = x …(i)

In △RNS,

∠NRS = ∠PRQ = x  (vertically opposite angles)

∠RNS = 90° (given)

⇒ ∠NSR + ∠RNS + ∠NRS = 180°

⇒ ∠NSR = 90° - x …(ii)

Now in Quadrilateral PTRS

∠PTS = 90° (given)

⇒ ∠TPR = ∠PQR + ∠PRQ = 2x (exterior angle to triangle PQR)

⇒ ∠PRS = 180° - ∠PRQ = 180° - x (QRS is a st. Line)

⇒ ∠PTS + ∠TPR + ∠PRS + ∠TSR = 360° (angles of a quad. = 360°)

⇒ 90° + 2x + 180° - x + ∠TSR = 360°

⇒ ∠TSR = 90° - x …(iii)

From (ii) and (iii)

∠TSR = ∠NSR

Therefore, QS bisects ∠TSN.

20. In the given figure, D and E are points on AB and AC respectively. AE and CD intersect at P such that AP = CP. If ∠BAE = BCD, prove that DBDE is isosceles.

Join DE and AC,

In △APD and △EPC,

∠DAP = ∠ECP ...(∵ ∠BAE = ∠BCD)

AP = CP …(given)

∠APD = ∠EPC …(vertically opposite angles)

∴ △APD ≅ △EPC …(By ASA Congruence criterion)

⇒ AD = EC …(c.p.c.t)

In △APC,

AP = CP …(given)

⇒ ∠PAC = ∠PCA  …(angles opposite to two equal sides are equal)

Now, ∠BAE = ∠BCD and ∠PAC = ∠PCA

⇒ ∠BAC = ∠BCA

⇒ BC = BA ...(sides opposite to two equal angles are equal)

⇒ BE + EC = BD + DA

⇒ BE = BD (∵ EC = DA)

⇒ ∠BDE = ∠BED …(angles opposite to two equal sides are equal)

⇒ △BDE is an isosceles triangle.

21. In DPQR as shown, ∠PQS = ∠RQS and QS ⊥ PR. Find the value of x and y, if PQ = 3x + 1; QR = 5y – 2; PS = x + 1 and SR = y + 2.

In △PQS and △SQR,

QS = QS …[common]

∠QSP = ∠QSR …[each = 90°]

∠PQS = ∠RQS …[given]

∴ △PQS ≅ △SQR …[By ASA criterion]

⇒ PS = RS

⇒ x + 1 = y + 2

⇒ x = y + 1 …(i)

And PQ = SQ

⇒ 3x + 1 = 5y – 2

⇒ 3(y + 1) + 1 = 5y – 2 …[From (i)]

⇒ 3y + 3 + 1 = 5y – 2

⇒ 2y = 6

⇒ y = 3

Putting y = 3 in (i),

x = y + 1 = 3 + 1 = 4

22. If PQS = 60°,

a. find QPR.

b. show that PQ = PS = QS = SR

(a) In △PQS,

PQ = QS …(given)

⇒ ∠QSP = ∠QPS …(angles opposite to two equal sides are equal)

Now, ∠PQS + ∠QSP + ∠QPS = 180°

⇒ 60° + 2∠QSP = 180°

⇒ 2∠QSP = 120 °

⇒ ∠QSP = 60°

⇒ QPS = 60°

In △PRS,

PS = SR …(given)

⇒ ∠PRS = ∠RPS …(angles opposite to two equal sides are equal)

By exterior angle property,

∠QSP = ∠RPS + ∠PRS

⇒ 60° = 2∠RPS

⇒ ∠RPS = 30°

Now ∠QPR = ∠QPS + ∠RPS = 60° + 30° = 90°

(b) In △PQS,

∠PQS = 60°, ∠QPS = 60° and ∠QSP = 60°

⇒ △PQS is an equilateral triangle.

⇒ PQ = QS = PS

And, PS = SR

⇒ PQ = PS = QS = SR

23. In the given figure, if DPQR is an isosceles triangle, prove that : QSR = exterior PRT.

Let ∠PQS = SQR = x and ∠PRS = ∠SRQ = y

In △PQR,

∠QPR + ∠PQR + ∠PRQ = 180°

⇒ ∠QPR + 2x + 2y = 180°

⇒ ∠QPR + 2x + 2y = 180°

⇒ ∠QPR = 180° - 2x – 2y …(i)

Since PQ = PR,

∠PRQ = ∠PQR …(angles opposite to two equal sides are equal)

⇒ 2x = 2y

⇒ x = y

Now, ∠PRT = ∠PQR + ∠QPR …(by exterior angle property)

⇒ ∠PRT = 2x + 180° - 2x – 2y …[From (i)]

⇒ ∠PRT = 180° - 2y …(ii)

In △SQR,

∠QSR + ∠SQR + ∠SRQ = 180

⇒ ∠QSR + x + y = 180°

⇒ ∠QSR = 180° - x – y

⇒ ∠QSR = 180° - y – y ….[∵ x = y (proved)]

⇒ ∠QSR = 180° - 2y …(iii)

From (ii) and (iii),

∠QSR = ∠PRT

24. In the given figure, if DABC is an isosceles triangle and PAC = 110°, find the base angle and vertex angle of the DABC.

Given : ∠PAC = 110°

To find:

Base angles : ∠ABC and ∠ACB

Vertex angle: ∠BAC

∠APQ + ∠PQC + ∠ACQ + ∠PAC = 360°

⇒ 90° + 90° + ∠QCA + 110° = 360°

⇒ ∠ACQ = 360° - 290°

⇒ ∠ACQ = 70°

⇒ ∠ACB = 70° …(i)

In △ABC,

AB = AC …(given)

⇒ ∠ACB = ∠ABC ...(angles opposite to two equal sides are equal)

⇒ ∠ABC = 70° …[From (i)]

In △ABC,

∠ABC + ∠ACB + ∠BAC = 180° …(angle sum property)

⇒ 70° + 70° + ∠BAC = 180°

⇒ ∠BAC = 180° - 140°

⇒ ∠BAC = 40°