Hi,
Could someone help me with this question? No matter what I do, I can't find the answer! I've posted on other forums, but no one has been able to figure it out.
Thanks!
Hello Satsue
Welcome to Math Help Forum!
I think the reason no-one has been able to prove it is that it just isn’t true! In order to prove the triangles similar, you need to show that their angles are equal. They have the angle P in common, so that’s no problem. Then you’d need to show that $\displaystyle \angle PYX =\angle PZY$. Since $\displaystyle \angle ZYX = \angle PZQ = 90^o$, we can do this provided we can show that $\displaystyle \angle QYZ = \angle QZY$. In other words, that the triangle ZQY is isosceles, with ZQ = YQ.
But it is perfectly possible to draw a diagram meeting all the given facts where ZQ is clearly not the same length as YQ – see the diagram I have attached.
Incidentally, notice that the line ZR and the point R are completely irrelevant to this question. This makes me think that there might be some additional fact that you haven’t told us about! Could you check to see whether you’ve missed something?
Grandad