http://img249.imageshack.us/img249/5...rclesf7.th.jpg

In the diagram, the circle ABC passes through the centre X of circle ABD, and CBD is a straight line. Prove that triangle ACD is isosceles.

Printable View

- April 24th 2007, 10:30 PMacc100jtNeed help on Circle properties
http://img249.imageshack.us/img249/5...rclesf7.th.jpg

In the diagram, the circle ABC passes through the centre X of circle ABD, and CBD is a straight line. Prove that triangle ACD is isosceles. - April 26th 2007, 09:18 AMSoroban
Hello, acc100jt!

I think I've got it . . .

Quote:

http://img249.imageshack.us/img249/5...rclesf7.th.jpg

In the diagram, the circle ABC passes through the centre X of circle ABD,

and CBD is a straight line.

Prove that triangle ACD is isosceles.

Draw radii XA, XB, and XD.

∆XAD is isosceles: .__/__XAD =__/__XDA = α

∆XBD is isoscles: .__/__XDB =__/__XBD = β . → .__/__CBX .= .180° - B

Hence: .__/__D .= .__/__XDA +__/__XDB .= .α + β

CAXB is a cyclic quadrilateral; its opposite angles are supplementary.

Since__/__CBX = 180° - β, then__/__CAX = β.

Hence: .__/__A .= .__/__XAD +__/__XAC .= .α + β

Therefore: .__/__A =__/__D, .CA = CD . → . ∆ACD is isosceles.