Question

Triangle ABC and Triangle DBC are two isosceles Triangle on the same base BC and Vertices. Aand D are on the same side of BC.If AD is extended to intersect BC at P. Show that Triangle ABD =~ Triangle ACD.



plese give me solution plese ​

Answer 1

Step-by-step explanation:

given-Triangle ABC and Triangle DBC are two isosceles Triangle-1

to prove-ABD =~ Triangle ACD.

proof-in Triangle ABD Triangle ACD

we have,

AB=AC(ABC is isosceles triangle)

DB=DC(DBC is isoaceles triangle)

AD=AD(common side)

by SSS CONGRUENCY

Triangle ABD =~ Triangle ACD.

Answer 2

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 ]

Other parts referred as attachment..