# circle chord and tangent

• Sep 23rd 2010, 08:39 AM
chrizzle
circle chord and tangent
PQ is a diameter of a circle and AB is a perpendicular chord cutting it at N. Prove that PN is equal in length to the perpendicular from P on to the tangent at A.

I have attached a diagram of how i imagine this situation to look. If my diagram is correct and I'm supposed to be proving that the two yellow segments are equal, then I'm not sure how to prove this.
However, if my diagram is wrong could someone please correct it so i can visulaise the situation.

Cheers
• Sep 23rd 2010, 08:53 AM
Traveller
What you have drawn is a perpendicular from the tangent to PQ. But what is required is a perpendicular from PQ to the tangent. Suppose it intersects the tangent at X, then angle PXA will be 90 degrees. In your diagram, if you draw it, a part of it will be inside the circle.
• Sep 23rd 2010, 09:10 AM
Unknown008
As the chord moves to the right, the tangent would meet the perpendicular to P in your drawing to nearly infinity, which does not agree with the proof.

I think this is how the drawing is meant to rather be:

http://p1cture.me/images/30570901435675219409.png

As for the proof, I haven't found out yet.

EDIT: That happened again! Sorry traveller, I didn't know you had posted before I do...
• Sep 23rd 2010, 09:44 AM
Unknown008
Okay, if we can prove that triangle PXA (I'll call the intersection between the tangent and the line from P as X) is similar to triangle PAN, then PX will automatically be equal to PN.

http://p1cture.me/images/23715696373296764241.png

Here, I call the angle theta, the angle APN. Angle PAO is also equal to theta. This makes angle XAP 90-theta.

Hence, XPA is equal to theta.

Since triangle XPA and triangle APN are similar and they share a common length AP, they are actually the same triangle, and PX = PN.