# Thread: [SOLVED] Application of intersecting chords theorem

1. ## [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?

2. Originally Posted by HelenaStage
Hi, I'm stuck halfway with this problem...
ABCD is a semi-circle with diameter AB. 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?
seems that you're proving AP.AC - DP.DB = AD^2 ?

you say you got
AP.AC - DP.DB = AP^2 - DP^2 which is correct, so why not finish it with the Pythagorean theorem?

3. But the sides I can prove don't form a right-angled triangle?

4. since AB is diameter, triangle ABD has right angle at D.. so APD is also right-angled triangle

5. 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...

6. 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