Un trapezio non è diverso

                Here we show and solve Problem 4, page 3 (for grade 7), submitted to a regional Mathematics Competition.

In translation, thanks to Mr DeepL:

               "Let ABCD be an isosceles trapezoid with $AB \parallel CD,\;AB>CD$. 

          We denote by $O$ the point of intersection of the diagonals of the trapezoid and let $M, N$ and $P$ be the middles of the line segments $[OC], [OB]$ and $[AD]$ respectively. Knowing $PM=MN$, find the measures of the angles of the triangle $PMN$."

ANSWER CiP

$60^{\circ}$ - the triangle $PMN$ is equilateral

 

 Solution CiP

          $MN$ is mid-segment/midline of triangle $OBC$ and we have $MN \overset{\parallel}{=}\frac{BC}{2}$. But $BC=AD$, so

 $$MN\;=\;\frac{AD}{2}. \tag{1}$$

          Because $PM=MN$, we also have

$$MP=\frac{AD}{2}. \tag{2}$$

But in the triangle $AMD$, the median $MP$ of side $AD$ is half that side. So $AMD$ is a right triangle with hypotenuse $AD$. 

       Since we have $DM \perp MA$, then in the triangle $CDO$(initially isosceles with $OC=OD$) the median $DM$ is also altitude, so it is still isosceles with $DO=DC$; ultimately it is equilateral triangle.

       Because in any trapezoid the triangles $AOB$ and $COD$ are similar, it follow that $ABO$ is equilateral triangle too. Then its median $AN$ is also its altitude, so we have $AN \perp ND$.

       Now we have the right triangle $ADN$ in which the median $NP$ of hypotenuse is

$$NP=\frac{AD}{2}.\tag{3}$$

       From relations (1), (2), (3) it follows that the triangle $PMN$ is equilateral, so it has all three angles equal to $60^{\circ}$.

$\blacksquare \;\;\;\blacksquare$