## A Problem for Dadu

- Raiyan Jamil
**Posts:**138**Joined:**Fri Mar 29, 2013 3:49 pm

### A Problem for Dadu

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.

A smile is the best way to get through a tough situation, even if it's a fake smile.

- Thanic Nur Samin
**Posts:**176**Joined:**Sun Dec 01, 2013 11:02 am

### Re: A Problem for Dadu

Solution:

Hammer with tact.

Because destroying everything mindlessly isn't cool enough.

Because destroying everything mindlessly isn't cool enough.

### Re: A Problem for Dadu

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.

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.