` A B C\ a n d\ D B C` are two isosceles triangles on the same bas `B C` and vertices `A\ a n d\ D` are on the same side of `B C` . If `A D` is extended to intersect `B C` at `P ,` show that ` A B D\ ~= A C D` (ii) ` A B P\ ~= A C P`

Explanation

(i) In `/_\ABD` and `/_\ACD`,
`AB=AC` [Since `/_\ABC` is isosceles]
`AD=AD` [common side]
`BD=DC` [Since `/_\BDC` is isosceles]
`/_\ABD~=/_\ACD` [By `SSS` criteria]
`:./_BAD=/_CAD`
`=>/_BAP=/_PAC` [By `CPCT`] ......(i)
Hence proved.

(ii) In `/_\ABP` and `/_\ACP`,
`AB=AC` [Since `/_\ABC` is isosceles]
`AP=AP` [common side]
`/_BAP=/_PAC` [from (i)]
`:./_\ABP~=/_\ACP` [By `SAS` criteria]
Hence proved.