∆ABC and ∆DBC are two isosceles I triangles on the same base BC an

1.	∆ABC and ∆DBC are two isosceles I triangles on the same base BC and vertices A and D are on the same side of BC (see figure). 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» (i) In ∆’s ABD and ACD, we have AB = AC (given) BD = DC (given) and AD = AD (common) ∴ ∆ABD ≅ ∆ACD (by SSS congruency rule) (ii) In ∆’s ABP and ACP, we have AB = AC (given) ∠BAP = ∠CAP and AP = AP (common) [∵ ∆ABD ≅ ∆ACD => ∠B AD = ∠CAD => ∠BAP = ∠CAP] ∴ ∆ABP ≅ ∆ACP (by SAS congruency rule) (iii) We have already proved in (i) that ∆ABD ≅ ∆CAD => ∠BAP = ∠CAP => AP bisects ∠A i.e. AP is the bisector of ∠A. In ∆’s BDP and CDP, we have BD = CD (given) BP = CP [∵ ∆ABP = ∆ACP] and DP = DP (common) ∴ ∆BDP ≅ ∆CDP (by SSS congruency rule) => ∠BDP = ∠CDP => DP is the bisector of ∠D. Hence, AP is the bisector of ∠A as well as ∠D. (iv) In (iii), we have proved that ∆BDP ≅ ∆CDP => BP = CP and ∠BPD = ∠CPD = 90°. ∴ ∠BPD and ∠CPD form a linear pair => DP is the perpendicular bisector of BC Hence, AP is the perpendicular bisector of BC.

∆ABC and ∆DBC are two isosceles I triangles on the same base BC and vertices A and D are on the same side of BC (see figure). 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»

(i) In ∆’s ABD and ACD, we have

AB = AC (given)

BD = DC (given)

and AD = AD (common)

∴ ∆ABD ≅ ∆ACD

(by SSS congruency rule)

(ii) In ∆’s ABP and ACP, we have

AB = AC (given)

∠BAP = ∠CAP

and AP = AP (common)

[∵ ∆ABD ≅ ∆ACD => ∠B AD = ∠CAD => ∠BAP = ∠CAP]

∴ ∆ABP ≅ ∆ACP

(by SAS congruency rule)

(iii) We have already proved in (i) that

∆ABD ≅ ∆CAD

=> ∠BAP = ∠CAP

=> AP bisects ∠A i.e. AP is the bisector of ∠A.

In ∆’s BDP and CDP, we have

BD = CD (given)

BP = CP [∵ ∆ABP = ∆ACP]

and DP = DP (common)

∴ ∆BDP ≅ ∆CDP

(by SSS congruency rule)

=> ∠BDP = ∠CDP
=> DP is the bisector of ∠D.

Hence, AP is the bisector of ∠A as well as ∠D.

(iv) In (iii), we have proved that

∆BDP ≅ ∆CDP

=> BP = CP and ∠BPD = ∠CPD = 90°.

∴ ∠BPD and ∠CPD form a linear pair

=> DP is the perpendicular bisector of BC

Hence, AP is the perpendicular bisector of BC.