We will investigate the congruency of the composite triangle BCD.
We are given that:
![<\text{CBF }\congUsing the sketch avaliable we can make the following statements:[tex]BF\text{ is a bisector of CD , imply If the point ( F ) is the midpoint of the line segment CD then we can say:[tex]CF\text{ }\cong FD]()
The line common to both the composite triangles is:
![BF\text{ is common}](https://img.qammunity.org/2023/formulas/mathematics/college/kiid13q4wo6h8ldmbb3lb3ys4t99kjha29.png)
Using the application of Pythagorean theorem we have:
![\begin{gathered} BC^2=CF^2+BF^2 \\ BD^2=FD^2+BF^2 \\ CF\text{ = FD} \\ =============== \\ BC^2=BD^2 \\ =============== \\ BC\cong BD \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/15wrvgt5ty4ecpa8f3twcs3k1vwktzu8g5.png)
Hence, using the congruency postulate for right angled triangles we have:
![\begin{gathered} \Delta CBF\text{ }\cong\Delta DBF \\ \text{RHS postulate} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/mydx3udw3knckzsnhyjl6c6opvs5gm5vw5.png)