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Given is acute triangle $ABC$. Let $D$ be foot of altitude from vertex $A$. Let $D_1$ be a point so that line of symmetry between $D_1$ and $D$ is line $AB$. Let $D_2$ be a point so that line of symmetry between $D_2$ and $D$ is line $AC$. Let points $E_1, E_2$ be on line $BC$ so that $D_1E_1 \parallel AB$ and $D_2E_2 \parallel AC$. Proof that points $D_1, E_1, D_2, E_2$ lie on same circle and that center of this circle lies on the circumscribed circle of triangle $ABC$. Sketch

My plan was to first prove that $D_1E_1E_2D_2$ is cyclic quadrilateral, in other words that $\angle E_1E_2D_2 + \angle E_1D_1D_2 = 180°$ or that $\angle E_1D_2D_1 = \angle E_1E_2D_1$. This would mean that points $D_1, E_1, D_2, E_2$ lie on same circle. However, without success.
Can someone help me with this please?

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  • $\begingroup$ Also , B and C are circumcenters. $\endgroup$ – Takahiro Waki Dec 3 '18 at 23:18
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put $\angle BAD=\angle EAC, \angle CAD=\angle ECB$.

EB is perpendicular $D_1E_1$. Since $BD_1E_1$ is isosceles triangle(since those angles are equal), $ED_1E_1$ is isosceles triangle, too. E is on two perpendiculars of $D_1E_1$ and $D_2E_2$. This shows that E is the circumcenter of these four points.

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  • $\begingroup$ Excellent answer, though it took me a while to clarify all the argument given. $\endgroup$ – Quang Hoang Dec 4 '18 at 4:33
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Let $ A' $ be the symmetric point of $ D $ about $ A $ on $ DA $, i.e $ A' $ is the point on $ DA $ such that $ A'A = DA $. Let $ F_1 $ be the intersection of $ DD_1 $ with $ AB $ and $ F_2 $ be the intersection of $ DD_2 $ with $ AC $. Then the following are easily observed:

(1) $ BF_1F_2C $ is cyclic. This is because $ AF_1DF_2 $ is cyclic as $ \angle AF_1D = \angle AF_2D = 90^{\circ} $. Then note that $ \angle BF_1F_2 = \angle BF_1D + \angle DF_1F_2 = 90^{\circ} + \angle DAF_2 = 90^{\circ} + 90^{\circ} - \angle C = 180^{\circ} - \angle BCF_2 $, proving the claim.

(2) The $ \triangle D_1A'D_2 $ is an expansion of $ \triangle F_1AF_2 $ with scaling factor $ 2 $. So $ A'D_1 $ (resp. $ A'D_2 $) is parallel to $ AB $ (resp $ AC $) and hence $ E_1 $ (resp $E_2 $) is the intersection of $ A'D_1 $ with $ BC $ (resp. $ A'D_2 $ with $ BC $). This shows that the quadrilateral $ D_1E_1E_2D_2 $ is obtained from the quadrilateral $ F_1BCF_2 $ by expansion about $ D $ by a factor of $ 2 $.

Since we've shown the cyclicity of $ F_1BCF_2 $ in (1), it follows that $ D_1E_1E_2D_2 $ is cyclic as well.

Haven't really thought of the circumcentrecentre as of now but should be tractable.

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    $\begingroup$ I think you have a typo in $(1)$. Shouldn't $\angle BF_1C$ be $\angle BF_1D$? $\endgroup$ – Snip3r Dec 5 '18 at 19:29
  • $\begingroup$ yes it is a typo, thank you for pointing out. $\endgroup$ – hellHound Dec 5 '18 at 19:30
  • $\begingroup$ Also, can you please explain a little bit more about why $\angle DF_1F_2 = \angle DAF_2$? $\endgroup$ – Snip3r Dec 5 '18 at 19:32
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    $\begingroup$ Because $ AF_1DF_2 $ is cyclic and these are angles subtended by the same chord. $\endgroup$ – hellHound Dec 5 '18 at 19:33

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