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Is it possible to prove that triangle CBF is congruent to Triangle DBF

Is it possible to prove that triangle CBF is congruent to Triangle DBF-example-1

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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}

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}

Hence, using the congruency postulate for right angled triangles we have:


\begin{gathered} \Delta CBF\text{ }\cong\Delta DBF \\ \text{RHS postulate} \end{gathered}

User Marek Dzikiewicz
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