Circle Geometry

• Oct 3rd 2010, 05:29 PM
Lukybear
Circle Geometry
ABCD is a cyclic quadrilateral. The diagonals AC and BD intersect at right angles at E. M is the midpoint of CD. ME produced meets AB at N.

Show that ME = MC

Sorry about not putting diagram. I have no idea at all at how to attempt this question
• Oct 3rd 2010, 08:04 PM
Educated
Drawing a diagram usually helps.

Is this what you are asking?
http://www.mathhelpforum.com/math-he...geometry-1.png
Prove that ME = MC.
Attachment 19170
• Oct 4th 2010, 01:08 AM
Lukybear
Attachment 19172

Sorry, wasnt access to a good computer in the morning. But it goes like this.
• Oct 5th 2010, 07:07 PM
Educated
Refer to my edited image:
http://www.mathhelpforum.com/math-he...y-1-solved.png
Attachment 19199

Here's my attempt at it:

We know that MC = MD because M is the midpoint of C and D.
Draw another circle, using point M as the centre, that touches point C and D.
Length CD is now the diameter of this new circle.
We know that triangle CED is a right angle triangle because it says that the 2 lines that form point E are at right angles.
For CED to be a right angle triangle, it MUST touch the circumference of the new circle. Why? Since the line CD is the diameter of the new circle, the only way for triangle CED to be right angled is if it touches the circumference of the new circle.

ME has the length of the radius of the new circle, and so does length MC.
Therefore, MC = ME.
• Oct 6th 2010, 03:22 AM
Lukybear
Yea thanks. I figured it out just before i came on. Thanks again tho.