ΔABC and ΔDBC are two isosceles triangles on the same base BC and ve

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.
Answer» (i) In ΔABD and ΔACD, AB = AC (Given) BD = CD (Given) AD = AD (Common) ∴ ΔABD ≅ ΔACD (By SSS congruence rule) ∴ ∠BAD = ∠CAD (By CPCT) ∴ ∠BAP = ∠CAP …. (1) (ii) In ΔABP and ΔACP, AB = AC (Given) ∠BAP = ∠CAP [From equation (1)] AP = AP (Common) ∴ ΔABP ≅ ΔACP (By SAS congruence rule) ∴ BP = CP (By CPCT) … (2) (iii) From equation (1), ∠BAP = ∠CAP Hence, AP bisects ∠A. In ΔBDP and ΔCDP, BD = CD (Given) DP = DP (Common) BP = CP [From equation (2)] ∴ ΔBDP ≅ ΔCDP (By S.S.S. Congruence rule) ∴ ∠BDP = ∠CDP (By CPCT) … (3) Hence, AP bisects ∠D. (iv) ΔBDP ≅ ΔCDP ∴ ∠BPD = ∠CPD (By CPCT) …. (4) ∠BPD + ∠CPD = 180 (Linear pair angles) ∠BPD + ∠BPD = 180 2∠BPD = 180 [From equation (4)] ∠BPD = 90^o … (5) From equations (2) and (5), it can be said that AP is the perpendicular bisector of BC.

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.

Answer»

(i) In ΔABD and ΔACD,

AB = AC (Given)

BD = CD (Given)

AD = AD (Common)

∴ ΔABD ≅ ΔACD (By SSS congruence rule)

∴ ∠BAD = ∠CAD (By CPCT)

∴ ∠BAP = ∠CAP …. (1)

(ii) In ΔABP and ΔACP,

AB = AC (Given)

∠BAP = ∠CAP [From equation (1)]

AP = AP (Common)

∴ ΔABP ≅ ΔACP (By SAS congruence rule)

∴ BP = CP (By CPCT) … (2)

(iii) From equation (1),

∠BAP = ∠CAP

Hence, AP bisects ∠A.

In ΔBDP and ΔCDP,

BD = CD (Given)

DP = DP (Common)

BP = CP [From equation (2)]

∴ ΔBDP ≅ ΔCDP (By S.S.S. Congruence rule) ∴

∠BDP = ∠CDP (By CPCT) … (3)

Hence, AP bisects ∠D.

(iv) ΔBDP ≅ ΔCDP

∴ ∠BPD = ∠CPD (By CPCT) …. (4)

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

∠BPD + ∠BPD = 180

2∠BPD = 180 [From equation (4)]

∠BPD = 90^o … (5)

From equations (2) and (5), it can be said that AP is the perpendicular bisector of BC.