∆ 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.
(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.
In Fig. 9.26, A, B, C and D are four points on a circle. AC and BD intersect at a point E such that ∠ BEC = 130° and ∠ ECD = 20°. Find ∠ BAC.