Find the angle $\angle MNP$ from the circle with centre C

Question

Let us consider a circle with centre at C, and diameter MCN. Consider a point P on the circumference of the circle and join MP and NP such that MPN is a triangle with $\angle MPN =π/2$. Now consider a point A on the same semicircle, I mean on the arc MP and two points B and D on the side MP such that ABCD is a square, recall that C is the centre of the circle. Then find the $\angle MNP$.

I mean I can not find a particular value of the required angle. If I assume radius as $r$, then somehow using triangles law I found a relation but not so close. Can you please help me to solve this question.

Thanks in advance.

Answer 1

Since $ABCD$ is a square and the diagonal $BD$ lies on the line segment $MP$, so does the midpoint of $BD$ (let's call it $O$), which is also the midpoint of $AC$.

We also have $AC\perp BD$. Hence $OC \perp MP$, and thus $MO = OP$.

Using $MC = CN$ and the midpoint theorem, we see that $OC$ is half the length of $PN$. But $OC$ is half the length of $AC$, which is the radius.

Hence $PN$ has equal length as the radius. This gives $\angle MNP = 60^\circ$.

Answer 2

Given $ABCD$ is a square, $AC \perp MP, CE = \frac{r}{2}, CE \parallel NP$

$\triangle MCE \sim \triangle MNP, PN = r$

So, $\angle MNP = 60^0$.