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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
- ΔABD ≅ ΔACD
- ΔABP ≅ ΔACP
- AP bisects ∠A as well as ∠D.
- AP is the perpendicular bisector of BC.
Solution
(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.
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