# Prove the following statement. Let ABC be any right triangle, the right angle at point C. The altitude drawn from C to the hypotenuse splits the triangle into two right triangles that are similar to each other and to the original triangle?

May 23, 2018

See Below.

#### Explanation:

According to the Question,

$\Delta A B C$ is a right triangle with $\angle C = {90}^{\circ}$, and $C D$ is the altitude to the hypotenuse $A B$.

Proof:

Let's Assume that $\angle A B C = {x}^{\circ}$.

So, $\angle B A C = {90}^{\circ} - {x}^{\circ} = {\left(90 - x\right)}^{\circ}$

Now, $C D$ perpendicular $A B$.

So, $\angle B D C = \angle A D C = {90}^{\circ}$.

In $\Delta C B D$,

$\angle B C D = {180}^{\circ} - \angle B D C - \angle C B D = {180}^{\circ} - {90}^{\circ} - {x}^{\circ} = {\left(90 - x\right)}^{\circ}$

Similarly, $\angle A C D = {x}^{\circ}$.

Now, In $\Delta B C D$ and $\Delta A C D$,

$\angle C B D = \angle A C D$

and $\angle B D C = \angle A D C$.

So, by AA Criteria of Similarity, $\Delta B C D \cong \Delta A C D$.

Similarly, We can find, $\Delta B C D \cong \Delta A B C$.

From that, $\Delta A C D \cong \Delta A B C$ .

Hope this helps.