A Problem for Dadu
- Raiyan Jamil
- Posts:138
- Joined:Fri Mar 29, 2013 3:49 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.
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.