# A geometry problem on power of points

An acute triangle $$ABC$$ is inscribed in a circumference of center $$O$$. Its heights are $$AD$$, $$BE$$ and $$CF$$. The line $$EF$$ intersects the circumference at two points, $$P$$ and $$Q$$.

(a) Prove that $$OA$$ is perpendicular to $$PQ$$.

(b) If $$M$$ is the midpoint of $$BC$$, prove that $$AP^2 = 2 \cdot AD \cdot OM$$

I can't get past the first part of the problem. It is sufficient to prove that $$\triangle APQ$$ is isosceles, which we can do by proving that $$\widehat{AP} = \widehat{AQ}$$. It is also sufficient to prove that $$AO$$ bissects $$PQ$$. Now the problem is in proving any of these equivalent statements.

I'm even more clueless on the second problem. I've tried assuming without proof that $$AO$$ is perpendicular to PQ and when on fiddling with the power of a point theorem, with no success.

I'd love a solution or at least some hints to this problem.

• geogebratube.org/student/m96341 if you download this document from geogebra , you can demonstrate (a) easily enough. ( I'm testing to see if you can download geogebra documents easily.)
– Alan
Apr 5 '14 at 21:39
• Click "on" Orthocenter , Circumcenter , and Circumcircle then you'll be able to create the construction very quickly.
– Alan
Apr 5 '14 at 22:43
• Actually, I can't demonstrate it easily enough. (I can't demonstrate it at all) I can see it is a $90^{\circ}$ angle very clearly, but I've tried to prove it in every way I could imagine, without success. Apr 6 '14 at 1:54
• One thing I've noticed about the symmetric nature of this construction is that I can find three new points on the circumcircle and join them together to form a triangle. which I have no idea what to call. It certainly isn't new !
– Alan
Apr 6 '14 at 3:53

Meant as hints if you read a line at a time, but otherwise a complete answer when you work through the details

1) consider the tangent at A, which is perpendicular to OA.
It suffices to show that the tangent is parallel to EF.
It suffices to show that angle AFE is equal to angle ACB.
This is well known, and can easily be shown because AFHE is a cyclic quad, so angle AFE is equal to angle AHE.

2) recall the Euler line. This shows us that 2OM=AH.