The following quadrilateral is a square also there are some known angles.prove that The segments of the inner triangle are equal.
My Attempt:If we name the inner point $O$ then two triangles $AOD$ and $BOC$ are equal then we proof that $OD=OC$ And how to prove the third segment is equal too?
$\qquad\qquad\qquad$
Let $E$ be a point inside $\triangle{OBC}$ such that $\triangle{EBC}$ is an isosceles triangle with $\angle{EBC}=\angle{ECB}=15^\circ$.
Then, since $\triangle{OEB}$ is an isosceles triangle with $OB=BE$ and $\angle{OBE}=60^\circ$, we know that $\triangle{OEB}$ is an equilateral triangle, and so $\triangle{OEC}$ is an isosceles triangle with $\angle{EOC}=(180^\circ-60^\circ-75^\circ-15^\circ)/2=15^\circ$.
Therefore, since $\angle{COB}=75^\circ$, we know that $\angle{DOC}=360^\circ -150^\circ-2\times 75^\circ=60^\circ$. With $OD=OC$, we can see that $\triangle{OCD}$ is an equilateral triangle.