Geometry Help

[matthew042]

Could you please help me on this problem? I really am stuck. I already found out that triangle EDZ is an isosceles triangle. Given that angle DZE = angle EZY, angle CYE = angle AYE, and ED//YZ, prove that CD = DZ - CY. Thanks everyone.

 

[Deleted member 4993]

You need prove triangle ECY is isosceles.

Since ED||AZ, and EY crosses those transversely,

mCEY = mAYE = mCYE ..............[edited: thanks, Bill]

Hence triangle ECY is isosceles.

 

[wjm11]

mCEY = mAEY = mCYE

small typo: mAEY should be mAYE

 

[Mrspi]

Could you please help me on this problem? I really am stuck. I already found out that triangle EDZ is an isosceles triangle. Given that angle DZE = angle EZY, angle CYE = angle AYE, and ED//YZ, prove that CD = DZ - CY. Thanks everyone.

[attachment=0:snu7f407]picture.jpg[/attachment:snu7f407]

This looked like a REALLY HARD problem at first.

But, it's not so bad.

CD + EC = ED (by the segment addition postulate...or segment addition, depending on your textbook)

If you looked at the other posts, you know that triangle ECY is isosceles, and why that's true. So, EC = CY.

You can substitute CY for EC, and you have

CD + CY = ED

And

CD = ED - CY

Since triangle EDZ is isosceles (from the given), you know that ED = DZ.

So, CD = DZ - CY

 

[mmm4444bot]

… by the segment addition postulate …

I envy people who can say things like this. :mrgreen: