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.