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Can you provide a proof for the following claim:

Claim. Given an arbitrary $\triangle ABC$. The $\triangle ADB$ , $\triangle BEC$ and $\triangle CFA$ are constructed on the sides of the $\triangle ABC$ externally , such that $\angle BAD=\angle ECB$ , $\angle DBA=\angle ACF$ and $\angle CBE=\angle FAC$ . Let $H_1$,$H_2$,$H_3$ be the orthocenters of the $\triangle ADB$ , $\triangle BEC$, $\triangle CFA$, respectively. Then the line segments $AH_2$, $BH_3$ and $CH_1$ concur at the point $P$.

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GeoGebra applet that demonstrates this problem can be found here.

My idea is to apply Jacobi's theorem on this problem. In order to do that we need to prove the following identities $\angle BAH_1=\angle H_3AC$, $\angle H_1BA=\angle CBH_2$ and $\angle H_2CB=\angle ACH_3$ but I don't know how this can be done. Any hints are welcomed.

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Clearly in $\triangle BDC$, $\angle HBE$ is complementary to $\angle ACB$.

Hence in the OP's diagram, $\angle H_3AC = 90 - \angle ACF = 90 - \angle DBA = \angle BAH_1$.

Same for others.

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