In acute $△ABC$, $AB>AC$. Let $M$ be the midpoint of $BC$. The exterior angle bisector of $\angle BAC$
meets ray $BC$ at $P$. Points $K$ and $F$ lie on line $PA$ such that $MF⊥BC$ and $MK⊥PA$. Prove that
$BC^2=4PF.AK$
IMO MOCK5 (V)
- Nadim Ul Abrar
- Posts:244
- Joined:Sat May 07, 2011 12:36 pm
- Location:B.A.R.D , kotbari , Comilla
$\frac{1}{0}$
- Nadim Ul Abrar
- Posts:244
- Joined:Sat May 07, 2011 12:36 pm
- Location:B.A.R.D , kotbari , Comilla
Re: IMO MOCK5 (V)
My solution :
Let the circumcircle of the triangle $ABC$ intersect the line $MF$ at the
$\frac{1}{0}$