Question

In triangle $ABC$, $BC = 4$, $AC = 3 \sqrt{2}$, and $\angle C = 45^\circ$. Altitudes $AD$, $BE$, and $CF$ intersect at the orthocenter $H$. Find $AH:HD$.
I tried using the pythagorean theorem to find side lengths, algebraically, but I couldn't find anything. Help!

Answer 1

I think this will help you out:

http://jwilson.coe.uga.edu/EMAT6680Fa2012/Szatkowski/SzatkowskiWU8/ASwriteup8.html

Answer 2

As in all geometry problems,
1. start with drawing a diagram indicating all given information.
2. deduce from known theorems additional information, and mark any new information on the diagram.
3. repeat #2 above.

I have done the first step for you, and the diagram is located at:
https://www.twiddla.com/fb5es1

Hints:
Note that congruent angles have been marked in read (step 2).

study the right-triangle BCE.
hence calculate mCE, then mEA (mark on diagram).
Study the right-triangle EAH, and deduce length of mAH (mark on diagram).

Similarly, study the right-triangle CAD.
Calculate mCD, hence mDB (mark on diagram).
Study right-triangle DBH, deduce length of mHD (mark on diagram).

Finally calculate ratio mAH:mHD.

Answer 3

stop cheating on Aops alcumus

Answer 4

Yes, I agree!