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Given ΔAEC≌ ΔAFC and angle B ≌ angle D. Prove ΔEBC ≌ ΔFDC

Given ΔAEC≌ ΔAFC and angle B ≌ angle D. Prove ΔEBC ≌ ΔFDC-example-1
Given ΔAEC≌ ΔAFC and angle B ≌ angle D. Prove ΔEBC ≌ ΔFDC-example-1
Given ΔAEC≌ ΔAFC and angle B ≌ angle D. Prove ΔEBC ≌ ΔFDC-example-2

1 Answer

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It has already been shown that angles AEC and AFC are congruent.

Then, the angle that is supplementary to AEC and the angle supplementary to AFC are also congruent.


\text{angle supp. to AEC }=180^(\circ)-AEC\cong180-AFC=\text{ angle supp. to AFC}

Also, it has already been shown that:


\begin{gathered} \text{angle supp. to AEC }=\text{ BEC} \\ \\ \text{angle supp. to AFC }=\text{DFC} \end{gathered}

Therefore:


\text{BEC}\cong\text{ DFC}

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