` 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 P` bisects `"\ "/_A` as well as `/_D` `A P` is the perpendicular bisector of `B C`

Explanation

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`]
In `/_\ABP` and `/_\ACP`,
`AB=AC` [Since `/_\ABC` is isosceles]
`AP=AP` [common side]
`/_BAP=/_PAC` [from (i)]
`:./_\ABP~=/_\ACP` [By `SAS` criteria]
`:.BP=PC` [By `CPCT`] ...(ii)
`/_APB=/_APC` [By `CPCT`]
`/_\ABD~=/_\△ACD`
`/_BAD=/_CAD` [From (i)]
`:.AD` bisects `/_A`
`=>AP` bisects `/_A` ...(iii)
In `/_\BDP` and `/_\CDP`,
`DP=DP` [common]
`BP=PC` [from (ii)]
`BD=CD` [since `/_\BDC` is isosceles]
`/_\BDP~=/_\CDP` [By `SSS` criteria]
`:./_BDP=/_CDP` [CPCT]
`:. DP` bisects `/_D`
`=>AP` bisects `/_D` ....(iv)
From equation (iii) and equation (iv), we get:
`AP` bisects `/_A` as well as `/_D`.
`/_APB+/_APC=180^@` [angles in linear pair]
Also, `/_APB=/_APC` [from (ii)]
`:./_APB=/_APC=180^@/2=90^@`
`BP=PC` and `/_APB=/_APC=90^@`
Hence, `AP` is perpendicular bisector of `BC`.