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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 \(YZ.\)

 Mar 15, 2022
 #2
avatar+117872 
+1

 

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

 

https://www.geogebra.org/classic/n39xzurk

 Mar 17, 2022
 #3
avatar+117872 
+1

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 = \(11\frac{3}{7}\)

 Mar 17, 2022
 #4
avatar+226 
+1

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.

 Mar 17, 2022
 #5
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Thanks Tiggsy,  (and it is nice to see you again  laugh )

 

Why is angle XWZ equal to  angle WYX ?   

Melody  Mar 17, 2022
 #6
avatar+226 
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Hi Melody.

They're both, for different reasons, equal to the angle between WX and the tangent at X.

 Mar 17, 2022
 #7
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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   laugh

https://www.youtube.com/watch?v=LCfgfg8Jv8g

 

Melody  Mar 17, 2022
edited by Melody  Mar 17, 2022
 #8
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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

 Mar 18, 2022
 #9
avatar+117872 
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Thanks Tiggsy,

That makes sense.  laugh

Melody  Mar 18, 2022

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