# Let $ABCD$ be a cyclic quadrilateral and let $AB$ and $CD$ meet at $E$. Let $M= (EBC)cap (EAD)$. Prove that $OMperp EM$

Mathematics Asked by Raheel on December 16, 2020

Let $$ABCD$$ be a cyclic quadrilateral and let $$AB$$ and $$CD$$ meet at $$E$$. Let $$M= (EBC)cap (EAD)$$. Prove that $$OMperp EM$$

I took midpoint of $$AB$$ as $$M_1$$ and midpoint of $$DC$$ as $$M_2$$ . I noticed that $$(EM_1OM_2)$$ is cyclic and it is enough to show that $$EM_1OM_2M$$ is cyclic.

PS: Diagram by @Shubhangi

## One Answer

You are very close!

Just note that M is the spiral center of the spiral similarity $$S$$ sending $$AB$$ to $$DC$$ . And hence the spiral similarity $$S$$ also take the midpoint of $$AB$$ to midpoint of $$DC$$.

So $$S:M_1 rightarrow M_2$$

So $$S:BM_1 rightarrow CM_2$$ .

So $$M$$ is the spiral center of the spiral symmetry which takes $$BM_1$$ to $$CM_2$$.

But notice that $$BM_1cap CM_2=E implies M =(EBC) cap (EM_1M_2)$$

So $$M in (EM_1M_2)$$ and by your observation, we get $$Min (EM_1OM_2)$$ , and hence we have $$OMperp EM$$.

Here M is called the miquel point and if we define $$F=BCcap DA$$ , then we have $$Min EF$$ if $$ABCD$$ is cyclic .

Correct answer by Sunaina Pati on December 16, 2020

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