How would you use given figure to justify the statement that the angle in a semicircle is $90^\circ$?

Let $BC$ is a diameter of the circle with centre $O$, and $A$ (in the figure called the apex) is a point on the circle forming triangle $ABC$ with $O$ the midpoint of the diameter.

In the figure, $OA = OB = r$, all equal to the radius.
So, the figure shows two isosceles triangles meeting at the apex.

Let the two base angles at the endpoints of the diameter be $a$ and $b$. By the isosceles triangle property:

The angle at the apex on the left-side isosceles triangle $= a$ (because that triangle has two equal sides $= r$).

The angle at the apex on the right-side isosceles triangle $= b$.

So, the total angle at the apex $= a + b$.

But in the triangle $ABC$ formed by the diameter and apex, the three angles sum to $180^\circ$:

$$\begin{aligned} a + b + (a + b) &= 180^\circ \\ \Rightarrow 2(a + b) &= 180^\circ \\ \Rightarrow a + b &= 90^\circ. \end{aligned}$$

So, the angle at the apex $= a + b = 90^\circ$. This justifies that the angle in a semicircle is $90^\circ$.