Question

"Question 1 ΔABC and ΔDBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see the given figure). If AD is extended to intersect BC at P, show that (i) ΔABD ≅ ΔACD (ii) ΔABP ≅ ΔACP (iii) AP bisects ∠A as well as ∠D. (iv) AP is the perpendicular bisector of BC.

Class 9 - Math - Triangles Page 128"

Answer 1

Congruence of triangles:
Two ∆’s are congruent if sides and angles of a triangle are equal to the corresponding sides and angles of the other ∆.

In Congruent Triangles corresponding parts are always equal and we write it in short CPCT i e, corresponding parts of Congruent Triangles.

It is necessary to write a correspondence of vertices correctly for writing the congruence of triangles in symbolic form.

Criteria for congruence of triangles:


SAS( side angle side):
Two Triangles are congruent if two sides and the included angle of a triangle are equal to the two sides and included angle of the the other triangle.


SSS(side side side):
Three sides of One triangle are equal to the three sides of another triangle then the two Triangles are congruent.

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Solution;

Given,
ΔABC and ΔDBC are two isosceles triangles in which AB=AC & BD=DC.

To Prove:

(i) ΔABD ≅ ΔACD
(ii) ΔABP ≅ ΔACP
(iii) AP bisects ∠A as well as ∠D.
(iv) AP is the perpendicular bisector of BC.


Proof:


(i) In ΔABD and ΔACD,
AD = AD (Common)
AB = AC (given)
BD = CD (given)
Therefore, ΔABD ≅ ΔACD (by SSS congruence rule)
∠BAD = ∠CAD (CPCT)
∠BAP = ∠CAP


(ii) In ΔABP & ΔACP,
AP = AP (Common)
∠BAP = ∠CAP
(Proved above)

AB = AC (given)

Therefore,
ΔABP ≅ ΔACP
(by SAS congruence rule).


(iii)
∠BAD = ∠CAD (proved in part i)

Hence, AP bisects ∠A.
also,

In ΔBPD and ΔCPD,
PD = PD (Common)
BD = CD (given)
BP = CP (ΔABP ≅ ΔACP so by CPCT.)

Therefore, ΔBPD ≅ ΔCPD (by SSS congruence rule.)

Thus,
∠BDP = ∠CDP( by CPCT.)

Hence, we can say that AP bisects ∠A as well as ∠D.


(iv)
∠BPD = ∠CPD
(by CPCT as ΔBPD ≅ ΔCPD)

& BP = CP (CPCT)
also,

∠BPD + ∠CPD = 180° (BC is a straight line.)
⇒ 2∠BPD = 180°
⇒ ∠BPD = 90°

Hence,
AP is the perpendicular bisector of BC.

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Hope this will help you......

Answer 2

Answer:

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