Circle passing through the centers of two others'

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rah4927
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Circle passing through the centers of two others'

Unread post by rah4927 » Thu Nov 10, 2016 1:03 pm

$1.$ Let $ ABC$ be an acute triangle. Point $ D$ lies on side $ BC$. Let $ O_B, O_C$ be the circumcenters of triangles $ ABD$ and $ ACD$, respectively. Suppose that the points $ B, C, O_B, O_C$ lies on a circle centered at $ X$. Let $ H$ be the orthocenter of triangle $ ABC$. Prove that $ \angle{DAX} = \angle{DAH}$.

tanmoy
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Re: Circle passing through the centers of two others'

Unread post by tanmoy » Sat Nov 12, 2016 3:23 pm

It is easy to see that $\triangle ABO_{B} \sim \triangle ACO_{C}$.
So,$A$ is the center of spiral similarity which takes $BO_{B}$ to $CO_{C}$.Then from a well known result,we get that $A$ is the Miquel Point of cyclic quadrilateral $BCO_{B}O_{C}$.

Now let $BC \cap O_{B}O_{C}=P$,$AH \cap BC=Q$ and $AD \cap O_{B}O_{C}=R$.Then $XA \perp AP$,$AD \perp O_{B}O_{C}$ and $AQ \perp BC$.Therefore,$A,R,Q,P$ are cyclic.So,

$\measuredangle DAH=\measuredangle EAQ=\measuredangle EPQ=\measuredangle APE$ (Since $PBO_{B}A$ is cyclic and $O_{B}A=O_{B}B$) $=\dfrac {\pi} {2}-\measuredangle EAP=\measuredangle XAE=\measuredangle XAD$.

Q.E.D
"Questions we can't answer are far better than answers we can't question"

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