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
∠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.
Answer
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°
⇒ 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.
Answer
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.
Answer
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.
Answer
∠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.
In △CAD and △CBE
CA = CB (Isosceles triangle)
∠CDA = ∠CEB = 90°
∠ACD = ∠BCE (common)
Therefore,
△CAD ≅ △CBA (AAS criteria)
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).
Answer
(i)
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)
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
∠BAD = ∠CAD (proved)
∆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 = 2∠2 and ∠4 = 2∠3. 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.
In ∆CAD and ∆CBE
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.
In ∆ADE and ∆BAC
AE = AC
AB = AD
∠BAD = ∠EAC
∠DAC = ∠DAC = DAC (common)
⇒ ∠BAC = ∠EAD = EAD
Therefore, ∆ADE ≅ ∆BAC (SAS criteria)
Hence, BC = DE
13. In the figure, ∠BCD = ∠ADC and ∠ACB = ∠BDA. Prove that AD = BC and ∠A = ∠B.
∠BCD = ∠ADC
∠ACB = ∠BDA
∠BCD + ∠ACB = ∠ADC + BDA
⇒ ∠ACD = ∠BDCACD = BDC
In ∆ACD and ∆BCD
∠ACD = ∠BDCACD = BDC
∠ADC = ∠BCD
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
⇒ AD = DC …(ii)
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
AD = PM
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.
Answer
AB = AC
AD = AD
∠B = ∠C
Therefore, ∆ABD ≅ ∆ADC (SSA criteria)
Hence, BD = DC
Thus, AD bisects BC
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.
Answer
Since AB = AC
∠ABC = ∠ACB
But ∠DBC = ∠DCB
⇒ ∠ABD = ∠ACD
Now in ∆ABD and ∆ADC
AB = AC
AD = AD
∠ABD = ∠ACD
Therefore, ∆ABD ≅ ∆ADC (SSA criteria)
Hence, ∠BAD = ∠CAD
Thus, AD bisects ∠BAC.
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.
Answer
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.
Answer
∠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.
(i) BG = CH
(ii) AG = AH
Answer
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.
Answer
∠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.
Answer
∠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)
Adding (i) and (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.
Answer
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
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
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
Answer
Given:
D is mid-point of BC
⇒ BD = DC
DE = AD
To prove:
a. ∆ABD ≅ ∆ECD
b. AB = EC
c. AB || EC
BD = DC (given)
∠ADB = ∠CDE (vertically opposite angles)
AD = DE (given)
∴ 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.
Answer
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.
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:
b. AB = BL
c. DC = AL/2
Answer
Given:
ABCD is a parallelogram, where BE = CE
To prove:
a. ∆DCE ≅ ∆LBE
b. AB = BL
c. DC = AL/2
∠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.
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°