91Ó°ÊÓ

$\mathrm{Δ}ABC$ is a right angled triangle, $∠B$ is of $90°$, and $BX$ is an altitude.

The sum of internal angles of a triangle is always $180°$.

Consider the following figure:

Sum of internal angles of a triangle is $180°$.

So,

$\begin{array}{c}∠p+∠q+90°=180°\\ ∠p+∠q=180°−90°\\ =90°\end{array}$

Angle $p$ and $q$ form a complementary pair, that is, one acute angle in right angle triangle will always be compliment of the other acute angle.

Now in $\mathrm{Δ}AXB$ and $\mathrm{Δ}BXC$,

$\begin{array}{l}∠AXB~∠BXC\text{ â¶Ä„â¶Ä„â¶Ä„}\left(Sincebothareright\text{ }angles\right)\\ ∠XBA~∠XCB\text{ â¶Ä„â¶Ä„â¶Ä„}(Since∠q=∠XCB)\\ ∠XAB~∠XBC\text{ â¶Ä„â¶Ä„â¶Ä„}(Since∠p=∠XAB)\end{array}$

Using $AAA$, the triangle$\mathrm{Δ}AXB~\mathrm{Δ}BXC$.

In $\mathrm{Δ}AXB$, $\mathrm{Δ}BXC$, and $\mathrm{Δ}ABC$all corresponding angles are congruent.

Therefore, by $AAA$criterion for similar triangles, $\mathrm{Δ}AXB~\mathrm{Δ}BXC~\mathrm{Δ}ABC$.