Question

∆ ABC and ∆ DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see Fig.). If AD is extended to intersect BC at P, show that

(i) ∆ABD ≅ ∆ACD

(ii) ∆ABP≅ ∆ACP

(iii) AP bisects ∠A as well as ∠D.

(iv) AP is the perpendicular bisector of BC.

Answer 1

(i) In ∆ABD and ∆ACD,

AB = AC (Given)

BD = CD (Given)

AD = AD (Common)

∠∆ABD ≅ ∠∆ACD (By SSS congruence rule)

∠BAD = ∠CAD (By CPCT) BAP = CAP …. (1)

(ii) In ∠∆ABP and ∠∆ACP,

AB = AC (Given)

∠BAP = ∠CAP [From equation (1)]

AP = AP (Common)

∠∆ABP ≅ ∠∆ACP (By SAS congruence rule)

BP = CP (By CPCT) … (2)

(iii) From equation (1),

∠BAP = ∠CAP

Hence, ∠AP bisects ∠A.

In ∠∆BDP and ∠∆CDP,

BD = CD (Given)

DP = DP (Common)

BP = CP [From equation (2)]

∠∆BDP ≅ ∠∆CDP (By S.S.S. Congruence rule)

∠BDP = ∠CDP (By CPCT) … (3)

Hence, AP bisects D.

(iv) ∠∆BDP ≅ ∠∆CDP

( ∠BPD = ∠CPD (By CPCT) …. (4)

∠BPD + ∠CPD = 180 (Linear pair angles)

∠BPD + ∠BPD = 180 BPD 2 = 180 [From equation (4)]

∠BPD = 90 … (5)

From equations (2) and (5),

it can be said that AP is the perpendicular bisector of BC.