Question

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

1. ΔABD ≅ ΔACD
2. ΔABP ≅ ΔACP
3. AP bisects ∠A as well as ∠D.
4. AP is the perpendicular bisector of BC.

Answer 1

(i) In ΔABD and ΔACD,

AB = AC        ...(Given)

BD = CD       ...(Given)

AD = AD      ...(Common)

∴ ΔABD ≅ ΔACD     ...(By SSS congruence rule)

⇒ ∠BAD = ∠CAD    ...(By Corresponding parts of congruent triangles)

⇒ ∠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 Corresponding parts of congruent triangles)   …(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 SSS Congruence rule)

⇒ ∠BDP = ∠CDP        ...(By Corresponding parts of congruent triangles)   …(3)

Hence, AP bisects ∠D.

(iv) ΔBDP ≅ ΔCDP

∴ ∠BPD = ∠CPD         ...(By Corresponding parts of congruent triangles)  …(4)

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

∠BPD + ∠BPD = 180°

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

∠BPD = 90°       …(5)

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