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If the bisector of the vertical angle of a triangle bisects the base of the triangle. then the triangle is isosceles. GIVEN : triangle A B C in which A D is the bisector of angle A meeting B C in `D` such that `B D=DC TO PROVE : ` triangleA B C` is an isosceles triangle.

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It seems that for proving the triangles to be congruent, we have `angle1=angle2, AD=AD` and `BD=DC` and so, triangles are congruent. But you see the ease of congruency. It is ASS , which is not the case of any congruency. So we will see this problem with a different view.
Construction : Produce AD to E such that AD=DE.
Proof : In `DeltaABD` and `DeltaECD`,
`:' {(BD=DC,"(Given as D is the mid-point)"),(angle3=angle4,(VOA)),(AD=DE,"(construction)"):}`
`:. DeltaABD cong DeltaECD` (SAS)
`:. AB=EC` (c.p.c.t) ...(1)
and `angle1=angle5` (c.p.c.t) ...(2)
But `angle1=angle2` (given) ...(3)
`:. angle2=angle5` [fro (2) and (3)]
`implies" "EC=AC` ...(4)
( `:'` sides opposite to equal angles are equal)
`:.` from (1) and (4), we have
`AB=AC`
`:. DeltaABC` is an Isosceles triangle. Hence Proved.

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