Cyclic pentagon with spiral similarity

Let $ABCD$ be a convex quadrilateral. Let $|\measuredangle ABC|$=$|\measuredangle ACD|$ and $|\measuredangle ACB|$=$|\measuredangle ADC|$. Let $O$ be the circumcenter of triangle $BCD$, distinct from $A$. Prove, that $|\measuredangle OAC|$ is a right angle.

I'm supposed to use spiral similarity with center $A$ and midpoints of sides $BC$ and $CD$ (let them be $M_A$ and $M_B$, respectively) to show that $AOM_ACM_B$ is a cyclic pentagon, but i don't know how to proceed with it. How do I do this?

Notice first of all that $OM_ACM_B$ all lie on the same circle of diameter $OC$, because $\angle OM_AC=\angle OM_BC=90°$.

Triangles $ABC$ and $ACD$ are similar by construction, and so are triangles $AM_AC$ and $AM_BD$ (equal angles at $C$ and $D$, enclosing sides proportional).

It follows that $\angle AM_AC=\angle AM_BD$ and, by difference, $\angle AM_AO=\angle AM_BO$. From that we obtain that points $AOM_AM_B$ are concyclic and they thus belong to the circle of diameter $OC$. The thesis then easily follows. I think you should recheck the qustion.

Take $\angle ABC=\angle ADC\implies \angle ACB=\angle ACD$

This makes $AB$ and $ACD$ congruent , which implies circumcentre o BCD lies on the line AC, i.e. $\angle OAC=2$right angles

• In that limiting case $A=O$. – Aretino Nov 30 '17 at 15:56