S is a circle lying inside circle T and touching it at A. P is a point on T; PQ and PR are the chords of T through P tangent to S, at X and Y respectively. Show that angle QAR = 2 x angle XAY.
(NB. P is not A.)




David

This is trivial.

[All geometry is trivial to Kerwin. - The Editor]

Using opposite angles in cyclic quadilateral are supplementary, angle QAR = pi - angle P.
By the alternate segment theorem , angle PXY = angle XAY = angle PYX. The result comes out immediately.

Kerwin

That may be 'trivial' to you Kerwin but it certainly isn't to me! (Although forgetting the Alternate Segment Theorem I suppose would create a problem for getting that solution.)

James.

How do you prove that opposite angles in a cyclic quad are supplementary?

Brad

Good question. I wish I knew...

Aha... Actually it's easy to prove. Let the four corners of the cyclic quad by A, B, C, D (in clockwise order) and the centre of the circle be O. Now O A=O B=O C=O D so the trianges O A B, O B C, O C D, O D A are isosceles. Therefore:

angle O A B = angle O B A [1]

angle O B C = angle O C B [2]

angle O C D = angle O D C [3]

angle O D A = angle O A D [4]

Define the angles on each side of equations [1]-[4] to be p, q, r, s respectively. Now using the fact that angles in a triangle add to p, we get:

angle A O B=p-2p

angle B O C=p-2q

angle C O D=p-2r

angle D O A=p-2s

These four angles must add to 2p. So:

2p = 4p-2(p+q+r+s)

p+q+r+s=p

So angle A D C + angle A B C=p.

So the two opposite angles in a cyclic quad add to p.

I assume that's what supplementary means...