An Isosceles triangle is a triangle that has two equal sides. Also, the two angles opposite to the two equal sides are equal. In other words, we can say that “An isosceles triangle is a triangle which has two congruent sides”.
Solution:
Given
Given: ∆ABC and ∆DCB are isosceles ∆on the same base BC.
To Prove
(i) ΔABD ≅ ΔACD
(ii) ΔABP ≅ ΔACP
(iii) AP bisects ∠A as well as ∠D.
(iv) AP is the perpendicular bisector of BC.
i) in ∆ABD and ∆ACB
AB=AC
BD=CD
AD=AD
∆ABD≅∆ACD ————-(By S.S.S congruency)
ii) in ∆ABP and ∆ACP
AB=AC
∠ BAP≅∠CAP [∆ABD≅∆ACD BY C.P.CT]
AP=AP ———[common]
∴[∆ABD≅∆ACD ———–[By S.A.S congruency ]
iii) [∆ABD≅∆ACD]
∠BAD=∠CAD
AD, bisects ∠A
AP, bisects ∠A
In ∆ BDP and ∆DPB
BD=CD —————(GIVEN)
DP=PC ———-[∆AB≅ ∆ACP C.P.C.T]
DP=DP ———–[common]
∴∆BDP≅∆CDP (S.S.S)
∠BDP=∠CDP (C.P.C.T)
DP bisects ∠D
AP bisects ∠D ——(ii)
From above equations, AP bisects ∠ A as well as ∠ D
iv) ∠ AP +∠APC =180°
∠APB=∠APC ————-[∆ABP≅∆ACP C.P.CT]
∠APB + ∠APC=180°
2 ∠ APB=180°
∠APB=180/2=90°
BP=PC
∴AP is ⊥ bisects of BC.
