Question

I need help with finishing this proof and with the problem after that

Answer 1

Consider the triangle ABD and triangle CBD.

`\begin{gathered} \angle ABD\cong\angle CBD\text{ (Angle Bisector)} \\ BD\cong BD\text{ (Common sides)} \\ \angle ADB\cong\angle CDB\text{ (Angle bisector)} \\ \Delta ABD\cong\Delta CBD\text{ (By ASA condition)} \end{gathered}`

Thus, triangle ABD is simillar to triangle CBD.