Question 1:
∆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 the given 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 ∆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°
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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