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?

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- Sep 5th 2009, 10:22 AMHelenaStage[SOLVED] Application of intersecting chords theorem
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? - Sep 5th 2009, 11:57 AMTaluivren
- Sep 5th 2009, 12:04 PMHelenaStage
But the sides I can prove don't form a right-angled triangle?

- Sep 5th 2009, 12:18 PMTaluivren
since AB is diameter, triangle ABD has right angle at D.. so APD is also right-angled triangle

- Sep 5th 2009, 12:25 PMHelenaStage
Sorry, from your reply I realize I have made a typo in the original question! I'm so sorry! The diameter is AD, not AB...I'm afraid this leaves me with no right angle...

- Sep 5th 2009, 01:40 PMTaluivren
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