# Frank Solutions for Chapter 11 Triangles and their Congruency Mathematics ICSE

### Exercise 11.1

1. In the given figure, Q: R = 1: 2. Find:

(a) Q

(b) R

Given,

∠Q: ∠R = 1: 2

Let us consider ∠Q = x°

∠R = 2x°

Now,

∠RPX = ∠Q + ∠R [by exterior angle property]

105° = x° + 2x°

⇒ 105° = 3x°

We get,

x° = 35°

Therefore,

∠Q = x° = 35° and

∠R = 2x° = 70°

2. The exterior angles, obtained on producing the side of a triangle both ways, are 100° and 120°. Find all the angles of the triangle.

∠ABP + ∠ABC = 180° (Linear pair)

100° + ∠ABC = 180°

⇒ ∠ABC = 180° – 100°

⇒ ∠ABC = 80°

∠ACQ + ∠ACB = 180° (Linear pair)

⇒ 120° + ∠ACB = 180°

⇒ ∠ACB = 180° – 120°

⇒ ∠ACB = 60°

Now,

In △ABC,

∠A + ∠B + ∠C = 180° (Angle sum property of a triangle)

∠A + 80° + 60° = 180°

∠A = 180° – 80° – 60°

We get,

∠A = 40°

Therefore, the angles of a triangle are 40°, 60° and 80°

3. Use the given figure to find the value of x in terms of y. Calculate x, if y = 15°.

(2x – y)° = (x + 5°) + (2y + 25)° (Exterior angle property)

⇒ 2x° – y° = x° + 5° + 2y° + 25°

⇒ 2x° – x° = 2y° + y° + 30°

⇒ x° = 3y° + 30°

When y = 15°,

We have,

x° = 3 × 15° + 30°

⇒ x° = 45° + 30°

We get,

x° = 75°

4. In a triangle PQR, P + Q = 130° and P + R = 120°. Calculate each angle of the triangle.

Given

In △PQR,

∠P + ∠Q = 130°

WKT

∠P + ∠Q = ∠PRY (Exterior angle property)

∠PRY = 130°

∠PRY + ∠R = 180° (Linear pair)

⇒ 130° + ∠R = 180°

⇒ ∠R = 180° – 130°

We get,

∠R = 50°

Also,

Given

∠P + ∠R = 120°

Now,

∠P + ∠R = ∠PQX (Exterior angle property)

∠PQX = 120°

∠PQX + ∠Q = 180° (Linear pair)

⇒ 120° + ∠Q = 180°

∠Q = 180° – 120°

We get,

∠Q = 60°

In △PQR,

∠P + ∠Q + ∠R = 180° (Angle sum property of a triangle)

⇒ ∠P + 60° + 50° = 180°

⇒ ∠P = 180° – 110°

We get,

∠P = 70°

Therefore, the angles of △PQR are,

∠P = 70°

∠Q = 60° and

∠R = 50°

5. The angles of a triangle are (x + 10)°, (x + 30)°and (x – 10)°. Find the value of ‘x’. Also, find the measure of each angle of the triangle.

For any triangle,

Sum of measures of all three angles = 180°

Hence,

We have,

(x + 10)° + (x + 30)° + (x – 10)° = 180°

⇒ x° + 10° + x° + 30° + x° – 10° = 180°

⇒ 3x° + 30° = 180°

⇒ 3x° = 180° – 30°

We get,

3x° = 150°

⇒ x° = 50°

Now,

(x + 10)° = (50 + 10)°

⇒ (x + 10)° = 60°

⇒ (x + 30)° = (50 + 30)°

⇒ (x + 30)° = 80°

⇒ (x – 10)° = (50 – 10)°

⇒ (x – 10)° = 40°

Therefore, the angles of a triangle are 60°, 80° and 40°

6. Use the given figure to find the value of y in terms of p, q and r

Here, SR is produced to meet PQ at point E

In △PSE,

∠P + ∠S + ∠PES = 180°  (Angle sum property of a triangle)

⇒ p° + y° + ∠PES = 180°

⇒ ∠PES = 180° – p° – y° …(1)

In △RQE,

∠R + ∠Q + ∠REQ = 180°  (Angle sum property of a triangle)

⇒ (180° – q°) + r° + ∠REQ = 180°

⇒ ∠REQ = 180° – (180° – q°) – r°

⇒ ∠REQ = q° – r° …(2)

Now,

∠PES + ∠REQ = 180°  (Linear pair)

(180°– p° – y°) + (q° – r°) = 180°  [from (1) and (2)]

⇒ -p° – y° + q° – r° = 0

⇒ –y° = -q° + p° + r°

We get,

y° = q° – p° – r°

7. In the figure given below, if RS is parallel to PQ, then find the value of y.

In △PQR,

∠P + ∠Q + ∠R = 180°  (angle sum property)

⇒ 4x° + 5x° + 9x° = 180°

⇒ 18x° = 180°

⇒ x = 10

∠P = 4x° = 4 × 10°

∠P = 40°

∠Q = 5x° = 5 × 10°

⇒ ∠Q = 50°

∠QPR = ∠PRS  (Alternate angles)

And,

∠QPR = 40°

∠PRS = 40°

By exterior angle property,

∠PQR + ∠QPR = ∠PRS + y°

⇒ 40° + 50° = 40° + y°

We get,

y = 50°

8. Use the given figure to show that: p + q + r = 360°

By exterior angle property,

∠p = ∠PQR + ∠PRQ

∠q = ∠QPR + ∠PRQ

∠r = ∠PQR + ∠QPR

Now,

∠p + ∠q + ∠r = ∠PQR + ∠PRQ + ∠QPR + ∠PRQ + ∠PQR + ∠QPR

On further calculation, we get,

∠p + ∠q + ∠r = 2 ∠PQR + 2∠PRQ + 2 ∠QPR

⇒ ∠p + ∠q + ∠r = 2 (∠PQR + ∠PRQ + ∠QPR)

⇒ ∠p + ∠q + ∠r = 2 × 180° (Angle Sum property: ∠PQR + ∠PRQ + ∠QPR = 180°)

We get,

∠p + ∠q + ∠r = 360°

Hence,

∠p + ∠q + ∠r = 360°

9. In ABC and PQR and, AB = PQ, BC = QR and CB and RQ are extended to X and Y respectively and ABX = PQY. Prove that ABC PQR.

In △ABC and △PQR

AB = PQ

BC = QR

∠ABX + ∠ABC = ∠PQY + ∠PQR = 180°

So,

∠ABX = ∠PQY

∠ABC = ∠PQR

Therefore,

△ABC ≅ △PQR (SAS criteria)

Hence, proved

10. In the figure, CPD = BPD and AD is the bisector of BAC. Prove that CAP BAP and CP = BP.

In △BAP and △CAP

∠BAP = ∠CAP (AD is the bisector of ∠BAC)

AP = AP

∠BPD + ∠BPA = ∠CPD + ∠CPA = 180°

We get,

∠BPD = ∠CPD

∠BPA = ∠CPA

Therefore,

△CAP ≅ △BAP (ASA criteria)

So,

CP = BP

Hence, proved

11. In the figure, BC = CE and 1 = 2. Prove that GCB DCE.

In △GCB and △DCE

∠1 + ∠GBC = ∠2 + ∠DEC = 180°

⇒ ∠1 = ∠2

∠GBC = ∠DEC

So,

BC = CE

∠GCB = ∠DCE (vertically opposite angles)

Therefore,

△GCB ≅ △DCE (ASA criteria)

Hence, proved.

12. In the figure, AB = EF, BC = DE, AB and FE are perpendicular on BE. Prove that ABD FEC

Given that,

In △ABD and △FEC

AB = FE and

BD = CE (∵ BC = DE; CD is common)

Therefore,

∠B = ∠E

△ABD ≅ △FEC (SAS criteria)

Hence, proved

13. In the figure, BM and DN are both perpendiculars on AC and BM = DN. Prove that AC bisects BD.

In △BMR and △DNR

BM = DN

∠BMR = ∠DNR = 90°

∠BRM = ∠DRN (vertically opposite angles)

Hence,

∠MBR = ∠NDR (Sum of angles of a triangle = 180°)

△BMR ≅ △DNR (ASA criteria)

Therefore,

BR = DR

So,

AC bisects BD

Hence, proved

14. AD and BE are altitudes of an isosceles triangle ABC with AC = BC. Prove that AE = BD.

CA = CB  (Isosceles triangle)

∠CDA = ∠CEB = 90°

∠ACD = ∠BCE (common)

Therefore,

Hence,

CE = CD

But,

CA = CB

AE + CE = BD + CD

AE = BD

Hence, proved

15. In ABC, X and Y are two points on AB and AC such that AX = AY. If AB = AC, prove that CX = BY.

In △ABC

AB = AC

AX = AY

BX = CY

In △BXC and △CYB

BX = CY

BC = BC

∠B = ∠C  (Angles opposite to equal sides are equal)

Therefore,

△BXC ≅ △CYB  (SAS criteria)

So,

CX = BY

Hence, proved

### Exercise 11.2

1. Which of the following pairs of triangles are congruent? Give reasons

(i) ∆ABC; (BC = 5 cm, AC = 6 cm, C = 80°);

∆XYZ; (XZ = 6 cm, XY = 5 cm, X = 70°).

(ii) ∆ABC; (AB = 8 cm, BC = 6 cm, B = 100°)

∆PQR; (PQ = 8 cm, RP = 5 cm, Q = 100°)

(iii) ∆ABC; (AB = 5 cm, BC = 7 cm, CA = 9 cm);

∆KLM; (KL = 7 cm, LM = 5 cm, KM = 9 cm).

(iv) ∆ABC; (B = 70°, BC = 6 cm, C = 50°);

∆XYZ; (Z = 60°, XY = 6 cm, X = 70°);

(v) ∆ABC; (B = 60°, BC = 6 cm, AB = 8 cm);

∆PQR; (Q = 60°, PQ = 6 cm, PR = 10 cm).

(i)

In ∆ABC and ∆XYZ

AC = XZ

BC = XY

The included angle ∠C = 80° is not equal to ∠X i.e. 70°.

Now, for ∆ABC to be congruent to ∆XYZ, AB should be equal to XY and YZ should be equal to BC. Then, ∠A = ∠C and ∠X = ∠Z. So, the measure of ∠B will not be equal to ∠Y.

(ii)

In ∆ABC and ∆PQR

AB = PQ

∠B = ∠Q

BC can be equal to QR or AC can be equal to RP

Therefore,

∆ABC can be congruent to ∆PQR.

(iii)

In ∆ABC and ∆KLM

AB = LM

BC = KL

AC = KM

Therefore,

∆ABC ≅ ∆KLM (SSS criteria)

(iv)

In ∆ABC and ∆XYZ

∠B = ∠X

BC = XY

Y = 180° - (70° + 60°) = 50°

∠C = ∠Y

Therefore,

∆ABC ≅ XYZ (ASA criteria)

(v)

In ∆ABC and ∆PQR

∠B = ∠Q

BC = PQ

By Pythagoras theorem,

2. In ∆ABC and ∆PQR and, AB = PQ, BC = QR and CB and RQ are extended to X and Y respectively and ABX = PQY. Prove that ∆ABC ∆PQR.

In ∆ABC and ∆PQR and

AB = PQ

BC = QR

∠ABX + ∠ABC = ∠PQY + ∠PQR = 180°

∠ABX = ∠PQY

⇒ ∠ABC = ∠PQR

Therefore,

∆ABC ≅ ∆PQR (SAS criteria)

3. In the figure, CPD = BPD and AD is the bisector of BAC. Prove that ∆CAP ∆BAP and CP = BP.

In ∆BAP and ∆CAP

∠BAP = ∠CAP  (AD is the bisector of ∠BAC)

AP = AP

∠BPD + ∠BPA = ∠CPD + ∠CPA = 180°

∠BPD = ∠CPD

⇒ ∠BPA = ∠CPA

Therefore,

∆CAP ≅ ∆BAP  (ASA criteria)

Hence, CP = BP.

4. In the figure, BC = CE and 1 = 2. Prove that ∆GCB ∆DCE.

In ∆GCB and ∆DCE and

∠1 + ∠GBC = ∠2 + ∠DEC = 180°

∠1 = ∠2

⇒ ∠GBC = ∠DEC

BC = CE

∠GCB = ∠DCE  (vertically opposite angles)

Therefore,

∆GCB ≅ ∆DCE  (ASA criteria)

5. In ∆ABC, AB = AC and the bisectors of angles B and C intersect at point O. Prove that BO = CO and the ray AO is the bisector of angle BAC.

In ∆ABC,

Since AB = AC

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

BO and CO are angle bisector of ∠B and ∠C respectively

Hence, ∠ABO = ∠OBC = ∠BCO = ∠ACO

Join AO to meet BC at D

In ∆ABO and ∆ACO and

AO = AO

AB = AC

∠C = ∠B

Therefore, ∆ABO ≅ ∆ACO  (SAS criteria)

Hence, ∠BAO = ∠CAO

⇒ AO bisects angle BAC

In ∆ABO and ∆ACO

And AB = AC

AO = AO

∆ABO ≅ ∆ACO  (SAS criteria)

Therefore,

BO = CO

6. In the figure, AB = EF, BC = DE, AB and FE are perpendiculars on BE. Prove that ∆ABD ∆FEC

In ∆ABD and ∆FEC

AB = FE

BD = CE  (BC = DE; CD is common)

∠B = ∠E

∆ABD ≅ ∆FEC  (SAS criteria)

7. In the figure, BM and DN are both perpendiculars on AC and BM = DN. Prove that AC bisects BD.

In ∆BMR and ∆DNR

BM = DN

∠BMR = ∠DNR = 90°

∠BRM = ∠DRN  (vertically opposite angles)

Hence, ∠MBR = ∠NDR (sum of angles of a triangle = 180°)

∆BMR ≅ ∆DNR (ASA criteria)

Therefore, BR = DR

So, AC bisects BD.

8. In ∆PQR, LM = MN, QM = MR and ML and MN are perpendiculars on PQ and PR respectively. Prove that PQ = PR.

In ∆QLM and ∆RNM

QM = MR

LM = MN

∠QLM = ∠RNM = 90°

Therefore, ∆QLM ≅ ∆RNM  (RHS criteria)

Hence, QL = RN …(i)

Join PM

In ∆PLM and ∆PNM and

PM = PM (common)

LM = MN

∠PLM = ∠PNM = 90°

Therefore, ∆PLM ≅ ∆PNM (RHS criteria)

Hence, PL = PN …(ii)

From (i) and (ii)

PQ = PR

9. In the figure, RT = TS, 1 = 22 and 4 = 23. Prove that ∆RBT SAT.

∠1 = 2∠2 and ∠4 = 2∠3

1 = 22 and 4 = 23∠1 = ∠4  (vertically opposite angles)

⇒ 2∠2 = 2∠3 or ∠2 = ∠3 …(i)

∠R = ∠S (since RT = TS and angle opposite to equal sides are equal)

⇒ ∠TRB = ∠TSA …(ii)

In ∆RBT and ∆SAT

RT = TS

∠TRB = ∠TSA

∠RTB = ∠STA (common)

Therefore, ∆RBT ≅ ∆SAT  (ASA criteria)

10. AD and BE are altitudes of an isosceles triangle ABC with AC = BC. Prove that AE = BD.

CA = CB  (Isosceles triangle)

∠CDA = ∠CEB = 90°

∠ACD = ∠BCE  (common)

Therefore, ∆CAD ≅ ∆CBE  (AAS criteria)

Hence, CE = CD

But, CA = CB

⇒ AE + CE = BD + CD

⇒ AE = BD

11. In ∆ABC, X and Y are two points on AB and AC such that AX = AY. If AB = AC, prove that CX = BY.

In ∆ABC

AB = AC

AX = AY

⇒ BX = CY

In ∆BXC and ∆CYB

BX = CY

BC = BC

∠B = ∠C = C (AB = AC and angles opposite to equal sides are equal)

Therefore, ∆BXC ≅ ∆CYB (SAS criteria)

Hence, CX = BY

12. In the figure, AC = AE, AB = AD and BAD = EAC. Prove that BC = DE. AE = AC

∠DAC = ∠DAC = DAC (common)

Therefore, ∆ADE ≅ ∆BAC (SAS criteria)

Hence, BC = DE

13. In the figure, BCD = ADC and ACB = BDA. Prove that AD = BC and A = B. ∠ACB = ∠BDA

∠BCD + ∠ACB = ∠ADC + BDA

⇒ ∠ACD = ∠BDCACD = BDC

In ∆ACD and ∆BCD

∠ACD = ∠BDCACD = BDC

ADC = BCD CD = CD

Therefore, ∆ACD ≅ ∆BCD (ASA criteria)

Hence, AD = BC and ∠A = ∠B.

14. In the figure, AP and BQ are perpendiculars to the line segment AB and AP = BQ. Prove that O is mid-point of the line segments AB and PQ.

Since AP and BQ are perpendiculars to the line segment AB, therefore AP and BQ are parallel to each other.

In ∆AOP and ∆BOQ

∠PAO = ∠QBO = 90°

∠APO = ∠BQO (alternate angles)

AP = BQ

Therefore, ∆AOP ≅ ∆BOQAOP BOQ (ASA criteria)

Hence, AO = OB and PO = OQ

Thus, O is the mid-point of line segments AB and PQ.

15. ∆ABC is isosceles with AB = AC. BD and CE are two medians of the triangle. Prove that BD = CE.

CE is median to AB

⇒ AE = BE …(i)

BD is median to AC

But AB = AC …(iii)

Therefore, from (i), (ii) and (iii)

BE = CD

In ∆BEC and ∆BDC

BE = CD

∠EBC = ∠DCB (angles opposite to equal sides are equal)

BC = BC (common)

Therefore, ∆BEC ≅ ∆BDC (SAS criteria)

Hence, BD = CE.

16. Sides AB, BC and the median AD of ∆ABC are equal to the two sides PQ, QR and the median PM of ∆PQR. Prove that ∆ABC ∆PQR.

In ∆ABC and ∆PQR

BC = QR

AD and PM are medians of BC and QR respectively

⇒ BD = DC = QM = MR

In ∆ABD and ∆PQM

AB = PQ

BD = QM

Therefore, ∆ABD ≅ ∆PQMABD PQM (SSS criteria)

Hence, ∠B = ∠Q

Now in ∆ABC and ∆PQR

AB = PQ

BC = QR

∠B = ∠Q

Therefore, ∆ABC ≅ ∆PQRABC PQR (SAS criteria)

17. Prove that in an isosceles triangle the altitude from the vertex will bisect the base.

AB = AC

∠B = ∠C

Therefore, ∆ABD ≅ ∆ADC (SSA criteria)

Hence, BD = DC

18. In ∆ABC, AB = AC. D is a point in the interior of the triangle such that DBC = DCB. Prove that AD bisects BAC of ∆ABC.

Since AB = AC

∠ABC = ∠ACB

But ∠DBC = ∠DCB

⇒ ∠ABD = ∠ACD

AB = AC

∠ABD = ∠ACD

Therefore, ∆ABD ≅ ∆ADC (SSA criteria)

19. O is any point in the ∆ABC such that the perpendicular drawn from O on AB and AC are equal. Prove that OA is the bisector of BAC.

In ∆POA and ∆QOA

∠OPA = ∠OQA = 90°

OP = OQ (given)

AO = AO

Therefore, ∆POA ≅ ∆QOA (SSA criteria)

Hence, ∠PAO = ∠QAO

Thus, OA bisects ∠BAC

20. In ∆ABC, AB = AC, BM and CN are perpendicular on AC and AB respectively. Prove that BM = CN.

In ∆BNC and ∆CMB

∠BNC = ∠CMB = 90°

∠NBC = ∠MCB (AB = AC)

BC = BC

Therefore, ∆BNC ≅ ∆CMB (AAS criteria)

Hence, BM = CN

21. ∆ABC is an isosceles triangle with AB = AC. GB and HC are perpendiculars drawn on BC.

Prove that

(i) BG = CH

(ii) AG = AH

In ∆ABC

AB = AC

∠ABC = ∠ACB (equal sides have equal angles opposite to them) ...(i)

∠GBC = ∠HCB = 90° …(ii)

Subtracting (i) from (ii)

∠GBA = ∠HCA …(iii)

In ∆GBA and ∆HCA

∠GBA = ∠HCA (from iii)

∠BAG = ∠CAH (vertically opposite angles)

BC = BC

Therefore, ∆GBA ≅ ∆HCA (ASA criteria)

Hence, BG = CH and AG = AH

22. In ∆ABC, AD is a median. The perpendiculars from B and C meet the line AD produced at X and Y. Prove that BX = CY.

In ∆BXD and ∆CYD

∠BXD = ∠CYD (90°)

∠XDB = ∠YDC (vertically opposite angles)

BD = DC (AD is median on BC)

Therefore, ∆BXD ≅ ∆CYD (AAS criteria)

Hence, BX = CY

23. Two right-angled triangles ABC and ADC have the same base AC. If BC = DC, prove that bisects ∠BCD.

∠BAC = ∠DAC (90°)

BC = DC

AC = AC (common)

Therefore, ∆ABC ≅ ∆ADC (SSA criteria)

Hence, ∠BCA = ∠DCA

Thus, AC bisects ∠BCD

24. PQRS is a quadrilateral and T and U are points on PS and RS respectively such that PQ = RQ, PQT = RQU and TQS = UQS. Prove that QT = QU.

∠PQT = ∠RQU …(i)

∠TQS = ∠UQS …(ii)

∠PQS = ∠RQS

In ∆PQS and ∆RQS

∠PQS = ∠RQS

PQ = RQ (given)

QS = QS (common)

Therefore, ∆PQS ≅ ∆RQS (SAS criteria)

Hence, ∠QPS = ∠QRS

Now in ∆PQT and ∆RQU

∠QPS = ∠QRS

PQ = RQ (given)

∠PQT = ∠RQU (given)

Therefore, ∆PQT ≅ ∆RQU (ASA criteria)

Hence, QT = QU.

25. A is any point in the angle PQR such that the perpendiculars drawn from A on PQ and QR are equal. Prove that AQP = AQR.

Given,

AM ⊥ PQ and AN ⊥ QR

AM = AN

In ∆AQM and ∆AQN,

AM = AN  (given)

AQ = AQ  (common)

∠AMQ = ∠ANQ  (Each 90°)

So, by RHS congruence, we have

∆AQM ≅ ∆AQN

⇒ ∠AQM = ∠AQN  (c.p.c.t)

⇒ ∠AQP = ∠AQR

26. In the given figure P is a midpoint of chord AB of the circle O. Prove that OP ⟂ AB.

Given:

In the figure, O is centre of the circle and AB is chord.

P is the mid-point of AB

⇒ AP = PB

To prove: OP ⊥ AB

Construction: join OA and OB

Proof:

In ∆OAP and ∆OBP

OA = OB  [radii of the same circle]

OP = OP  [common]

AP = PB  [given]

∴ By Side - Side –Side criterion of congruency,

∆OAP ≅ ∆OBP

The corresponding parts of the congruent triangles are congruent.

∴ ∠OPA = ∠OPB

But ∠OPA + ∠OPB = 180°  [linear pair]

∴ ∠OPA = ∠OPB = 90°

Hence OP ⊥ AB.

27. In a circle with centre O. If OM is perpendicular to PQ, prove that PM = QM.

Given:

In the figure, O is centre of the circle and PQ is a chord.

OM ⊥ PQ

To prove: PM = QM

Construction: Join OP and OQ

Proof:

In right triangles ∆OPM and ∆OQM,

OP = OQ  [radii of the same circle]

OM = OM  [common]

∴ By right Angle-Hypotenuse-Side criterion of congruency,

∆ OPM ≅ ∆OQM

The corresponding parts of the congruent triangles are congruent.

∴ PM = QM

28. In a triangle ABC, if D is midpoint of BC; AD is produced upto E such as DE = AD, then prove that:

a. DABD and DECD are congruent.

b. AB = EC

c. AB is parallel to EC

Given:

D is mid-point of BC

⇒ BD = DC

To prove:

a. ∆ABD ≅ ∆ECD

b. AB = EC

c. AB || EC

a. In ∆ABD and ∆ECD,

BD = DC  (given)

∠ADB = ∠CDE  (vertically opposite angles)

∴ By Side-Angle-Side criterion of congruence,

∆ABD ≅ ∆ECD

b. The corresponding parts of the congruent triangles are congruent.

∴ AB = EC

c. Also, ∠DAB = ∠DEC  (c.p.c.t)

∴ AB || EC  (∠DAB and ∠DEC are alternate angles)

29. If the perpendicular bisector of the sides of a triangle PQR meet at I, then prove that the line joining from P,Q,R to I are equal.

Given :

In ∆PQR,

PA is the perpendicular bisector of QR ⇒ QA = RA

RC is the perpendicular bisector of PQ ⇒ PC = QC

QB is the perpendicular bisector of PR ⇒ PB = RB

PA, RC and QB meet at I.

To prove: IP = IQ = IR

Proof: In ∆QIA and ∆RIA

QA = RA  [Given]

∠QAI = ∠RAI  [Each = 90°]

IA = IA  [common]

∴ By Side-Angle-Side criterion of congruence,

∆QIA ≅ ∆RIA

The corresponding parts of the congruent triangles are congruent.

∴ IQ = IR …(i)

Similarly, in ∆RIB and ∆PIB

RB = PB  …[Given]

∠RBI = ∠PBI  ….[Each = 90°]

IB = IB  ….[common]

∴ By Side-Angle-Side criterion of congruence,

∆RIB ≅ ∆PIB

The corresponding parts of the congruent triangles are congruent.

∴ IR = IP  ….(ii)

From (i) and (ii), we have

30. In the given figure ABCD is a parallelogram, AB is produced to L and E is a midpoint of BC. Show that:

a. DDCE DLBE

b. AB = BL

c. DC = AL/2

Given:

ABCD is a parallelogram, where BE = CE

To prove:

a. ∆DCE ≅ ∆LBE

b. AB = BL

c. DC = AL/2

a. In ∆DCE and ∆LBE

∠DCE = ∠EBL   [DC || AB, alternate angles]

CE = BE  [given]

∠DEC = ∠LEB  [vertically opposite angles]

∴ By Angle-Side-Angle criterion of congruence,

∆DCE ≅ ∆LBE

The corresponding parts of the congruent triangles are congruent.

∴ DC = LB  …(1)

b. DC = AB  …(2) [Opposite sides of a parallelogram]

From (1) and (2),

AB = BL  ….(3)

c. AL = AB + BL

⇒ AL = AB + AB  [From (3)]

⇒ AL = 2AB

⇒ AL = 2DC  [From (2)]

32. In the given figure, AB = DB and AC = DC. Find the values of x and y.

In ∆ABC and ∆DBC

AB = DB  [given]

AC = DC  [given]

BC = BC  [common]

∴ By Side-Side-Side criterion of congruence,

∆ABC ≅ ∆DBC

∴ ∠ACB = ∠DCB [c.p.c.t]

⇒ y + 15° = 63°

⇒ y = 63° - 15°

⇒ y = 48°

Now, ∠ABC = ∠DBC  [c.p.c.t]

⇒ 29° = 2x - 4°

⇒ 2x = 29° + 4°

⇒ 2x = 33°

⇒ x = 33°/2

⇒ x = 16.5° and y = 48°