Question

In the given figure, $AB\perp BC,FG\perp BC$ and $DE\perp AC$. Prove that $△ADE$ ~ $△GCF$.

Answer 1

In the given figure we have to prove that $\Delta$ ADE and $\Delta$ FGC are similar.

Now from $\Delta$ ADE and $\Delta$ GFC we have,

$\angle AED=\angle GFC$ [right-angles],

$\angle DAE=\angle FGC$ [ $\because$ $AD\parallel GF$ and$AC$ intersector, similar angles].

Then $\angle ADE=\angle GCF$ [Remaining angles equal by ${180}^{\circ }$ rule].

$\therefore \Delta ADE$ is similar to $\Delta GFC.$[By AA test] $\left[henceproved\right]$