On the same base BC there are two isosceles triangles. Triangle ABC and Triangle DBC and are the opposite sides of BC. Show that Triangle ADC similar to Triangle ADB.
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To Show :
∆ABD ≅ ∆ACD.
∆ABP ≅ △ACP
AP bisects ∠A as well as ∠D
AP is perpendicular bisector of BC .
Proof :
In △ABD AND △ACD
AB = AC [ GIVEN]
AD = AD [ COMMON ]
BD = CD [ GIVEN ]
Therefore, ∆ABD ≅ ∆ACD.[SSS]
2. In △ABP and △ ACP
AB = AC [ GIVEN ]
∠BAP = ∠CAP [ C.P.C.T ]
AP = AP [ COMMON ]
Therefore , △ABP ≅ △ACP [ SAS ]
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