Hi, I'm stuck halfway with this problem...
ABCD is a semi-circle with diameter AD. P is the point of intersection of AC and BD. Prove that: AP.AC + DP.DB = AD^2
I can prove that
AP.AC - DP.DB = AP^2 - DP^2
Help?
consider AC intersects BD outside the circle (swap B and C on your picture). then we have:
AP.AC + DP.DB = AP.(AP - PC) + DP.DB = AP^2 - AP.PC + DP.DB = [now we use secants intersection theorem AP.PC = DP.BP] =
= AP^2 - DP.BP + DP.DB = AP^2 + DP.(DB - BP) = AP^2 + (DB + BP).(DB - BP) = AP^2 + DB^2 - BP^2 = (AP^2 - BP^2) + DB^2 =
= AB^2 + DB^2 = AD^2
last two equalities are pythagorean theorem.
If the situation is like on your picture with P inside the circle, calculation will be very similar (with use of chords intersection theorem this time, but this theorem is in fact the same as secants intersection theorem), try it!
AP.AC + DP.DB = AP.(AP + PC) + DP.DB = AP^2 + AP.PC + DP.DB = [now we use chords intersection theorem AP.PC = DP.BP] =
= AP^2 + DP.BP + DP.DB = AP^2 + DP.(DB + BP) = AP^2 + (DB - BP).(DB + BP) = AP^2 + DB^2 - BP^2 = (AP^2 - BP^2) + DB^2 =
= AB^2 + DB^2 = AD^2