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RHSStands for Right Angle-Hypotenuse-Side

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.

Solution 1:
(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)