Let in triangle ABC, D be the midpoint of BC and P be a point on AD such that

$\displaystyle \angle BAC + \angle BPC = 180^0$

Prove that $\displaystyle AC.BP = AB.CP$

Printable View

- Oct 22nd 2008, 05:46 AMalexmahonetriangle ABC
Let in triangle ABC, D be the midpoint of BC and P be a point on AD such that

$\displaystyle \angle BAC + \angle BPC = 180^0$

Prove that $\displaystyle AC.BP = AB.CP$