In the given figure, $ riangle FEC \cong riangle GDB$ and $\angle 1 = \angle 2$. Prove that $ riangle ADE \sim riangle ABC$.

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Question

In the given figure, $	riangle FEC \cong 	riangle GDB$ and $\angle 1 = \angle 2$. Prove that $	riangle ADE \sim 	riangle ABC$.

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Accepted Answer

Step 1: $	riangle FEC \cong 	riangle GDB \Rightarrow \angle 3 = \angle 4$ (CPCT)
Step 2: In $	riangle ABC$, $\angle 3 = \angle 4 \Rightarrow AB = AC$ ...(i)
Step 3: In $	riangle ADE$, $\angle 1 = \angle 2 \Rightarrow AD = AE$ ...(ii)
Step 4: Dividing (ii) by (i): $\frac{AD}{AB} = \frac{AE}{AC} \Rightarrow DE \parallel BC$ (by converse of BPT)
Step 5: $\angle 1 = \angle 3$ and $\angle 2 = \angle 4$ (corresponding angles). $	herefore 	riangle ADE \sim 	riangle ABC$ (AA similarity)

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