In the figure,if $\Delta ABE \cong \Delta ACD,$ show that $\Delta ADE \sim \Delta ABC.$
(N/A) Given: $\Delta ABE \cong \Delta ACD.$
Since the triangles are congruent,their corresponding parts are equal $(CPCT)$.
Therefore,$AB = AC$ $...(1)$
And,$AE = AD$ $...(2)$
Now,consider $\Delta ADE$ and $\Delta ABC$:
From equations $(1)$ and $(2)$,we have:
$\frac{AD}{AB} = \frac{AE}{AC}$
Also,$\angle DAE = \angle BAC$ (Common angle).
Therefore,by the $SAS$ (Side-Angle-Side) similarity criterion,$\Delta ADE \sim \Delta ABC.$
Since the triangles are congruent,their corresponding parts are equal $(CPCT)$.
Therefore,$AB = AC$ $...(1)$
And,$AE = AD$ $...(2)$
Now,consider $\Delta ADE$ and $\Delta ABC$:
From equations $(1)$ and $(2)$,we have:
$\frac{AD}{AB} = \frac{AE}{AC}$
Also,$\angle DAE = \angle BAC$ (Common angle).
Therefore,by the $SAS$ (Side-Angle-Side) similarity criterion,$\Delta ADE \sim \Delta ABC.$
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