A Problem for Dadu

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Raiyan Jamil
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A Problem for Dadu

Unread post by Raiyan Jamil » Mon Feb 20, 2017 4:16 pm

Let $ABCD$ be a cyclic quadrilateral. Let $H_A, H_B, H_C, H_D$ denote the orthocenters of triangles $BCD, CDA, DAB$ and $ABC$ respectively. Prove that $AH_A, BH_B, CH_C$ and $DH_D$ concur.
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Thanic Nur Samin
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Re: A Problem for Dadu

Unread post by Thanic Nur Samin » Mon Feb 20, 2017 6:09 pm

Solution:
Let the circumcircle of $ABCD$ be the unit circle. We will apply complex numbers. Now, let $a,b,c,d,h_a,h_b,h_c,h_d$ denote $A,B,C,D,H_A,H_B,H_C,H_D$ respectively. Now,

$h_a=b+c+d$

So, the midpoint of $AH_A$ has complex coordinate $\dfrac{a+h_a}{2}=\dfrac{a+b+c+d}{2}$.

Due to the symmetric nature of the midpoints, they all share a common midpoint. So, they are concurrent.
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rah4927
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Re: A Problem for Dadu

Unread post by rah4927 » Mon Feb 20, 2017 10:17 pm

We will show that the four lines are concurrent at their midpoints.

Lemma : $AH_AH_BB$ is a parallelogram

Proof : Since $AH_A=2\times \text{distance of O to CD}=BH_B$, and $AH_A||BH_B$, we are done.

Now it easily follows that the diagonals $AH_B$ and $BH_A$ bisect each other. We can do this for every pair of the four lines. Therefore, they all share a common midpoint.

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