Geometry proof.?

Question

This is going to sound very confusing, because I don't have a picture to show. I'll try my best to explain.

Okay, so I have a picture of a circle with the center point O and diameter AOH. There are 5 chords that are inside of the circle. One of them, labeled MT, crosses the circle, intersecting the diameter AOH. This makes an "X" shape almost, but MT does not cross through the middle of the circle. The other 4 chords outline connect the tips of the "X" shape. So the tips of the X are on the circumference of the circle, and are labeled M, A, T, H clockwise. O is the midpoint of the circle. Z is where chord MT hits the diameter AOH.

I have to prove that
(MZ)/(ZH) = (AZ)/(ZT)

I don't remember the rules to prove this. If you don't understand my explanation of the picture, and simple rules to prove things about circles would be very helpful. Thanks!

To Geezah:
No, there are a total of 6 chords.
Four points lay on the circumference of the circle. If you connect the points to make an uneven box, these are 4 of the chords (MA, AT, TH, and HM).
Then there is the diameter, another chord, AOH. Then another chord MT intersects the diameter at point Z.
I hope that makes more sense :/

To Lizzie:
There was no other given info, sadly. The only thing I can come up with is that segment HO is congruent to segment OA, because point O splits the diameter in half :/

Answer 1

All you need to do is show that triangle ZMH is similar to triangle AZT.

This is done by showing that MZH = angle AZT because vertical angles are equal, and angle HMT = angle HAT because they are inscribed angles intercepting the same arc AT.

Since these two triangles are similar, their corresponding sides are proportional.

Therefore MZ/ZH = AZ/ZT.

Answer 2

I was able to follow the directions and make the drawing, but I was wondering if there was any other information given. I know segment HA is the diameter and MATH is a quadrilateral and O is the center. There are statements about chords and angles inscribed in circles and semicircles that can be used, but I would think that there would be some other given info too.

Answer 3

I think you mean that there are only 2 chords, MT and AH, which intersect at some point Z that isn't the center of the circle. A chord by definition connects two points of a circle.

Here's how Euclid did it some 2300 years ago:
http://aleph0.clarku.edu/~djoyce/java/elements/bookIII/propIII35.html

Here's another proof that's a little easier to read:
http://www.cut-the-knot.org/proofs/IntersectingChordsTheorem.shtml

Answer 4

I just happened to ask an almost identical question yesterday. The proof can be found inthe site below.

It involves similar triangles. Since angles AHM and ATM are subtended on the same chord AM, they are equal. The same is true for HAT and HMT since they are both subtended on the same chord HT. You can also see that MZH = AZT.

There MZH and AZT are similar triangles.