Triangles `A B C` and +DBC are on the same base `B C` with A, D on opposite side of line `B C ,` such that `a r(_|_ A B C)=a r( D B C)dot` Show that `B C` bisects `A Ddot`

Explanation

Given: Triangles `ABC` and `DBC` are on the same base `BC` with `A`, `D` on opposite side of line `BC`, such that `ar(/_\ABC)=ar(/_\DBC)`.
To prove: `BC` bisects `AD`
Construction: Draw `AL_|_BC` and `DM⊥BC`.
Proof : `ar(/_\ABC)=ar(/_\DBC)` [Given]
`=>BC\xx\(AL)/2=BC\xx\(DM)/2`
`=>AL=DM` ....(i)
Now, In `/_\OAL` and `∆OMD`
`AL=DM` [From (i)]
`=>/_ALO=/_DMO` [Each`=90^@`]
`=>/_AOL=/_MOD` [Vert. opp. `/_s`]
`=>/_OAL=/_ODM` [Third angles of the triangles]
`:. /_\OAL=~/_\OMD` [By `ASA`]
`:.OA=OD` [By `CPCT`]
`therefore BC` bisects `AD`