The tangent to the circumcircle of triangle \(WXY\) at \(X\) is drawn, and the line through \(W\) that is parallel to this tangent intersects \(\overline{XY}\) at \(Z.\) If \(XY=14\) and \(WX = 6\) find \(YZ.\)
+118687
Here is a geometric solution. I just wanted to see if the distance actually did always stay the same.
The graph is interactive. You can right click W or Y and change the triangle.
The length of ZY stays the same. Approx 11.43 units
+118687 this is not a general solution. BUT
If you let the diameter of the circle be 14 units then WZ will be an altitufde for WXY
By using pythagoras's theorem and a bit of manipulation you can determine that ZX = \(11\frac{3}{7}\)
+397 This is a question that's appeared several times on this site, maybe with different numbers but basically the same question - just search circumcircle.
It drops out in a couple of lines if you notice that the triangles WXY and ZXW are similar.
That gets you ZX/6 = 6/14, so ZX = 36/14 = 18/7.
Then, ZY = 14 - 18/7 = 80/7.
+397 Hi Melody.
They're both, for different reasons, equal to the angle between WX and the tangent at X.
+118687 Thanks Tiggy.
I assume the accpetable reason is ...
"The angle between a tangent and a chord is equal to the angle subtended by the chord in the alternate segment."
But I do not remember this rule. I have to find/see a proof to convince myself that it is true.
I found a good proof on YouTube 
https://www.youtube.com/watch?v=LCfgfg8Jv8g
Hi Melody
I've taken a look at that youtube link and I think that they've chosen a really dreadful proof, it's so much easier than that.
Here's the way I look at it.
Using their diagram, chord AC, tangent at A, start by drawing the diameter of the circle at A meeting the circle again at D.
Triangle ACD will be a right angled, with the right angle at C and so the angle CDA will equal the angle between the chord and the tangent, (a couple of subtractions from 90 deg).
Now, since angles on the same arc are equal, angle ABC will equal angle angle CDA for any B in the same segment of the circle as D.
Tiggsy
