1. ## Circle Geometry Help Please

Hello. Could somebody please help me out on a little circle geometry? Normally, I'm okay at geometry but this one had me stumped so perhaps you mathematicians could shed a little light for me?

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A circle is drawn through the vertices of triangle ABC.
The perpendiculars CQ and AP are produced to meet the circumference of the circle at K and L (respectively).

Prove that angle LBP = angle ZBP

Here's a little diagram I drew on MS Paint:

Now, with no sides given, I assume I would have to show the triangle LBZ is isosceles by showing that the base angles are equal. This is what I have done so far:

I have managed to show that triangles KAZ and ZCL are isosceles (not sure if it helps).

I did these by letting angle ZKA be β.
Using 'angles subtended by the same arc' theorem, angle ABP is also β.
I used this theorem again to get angle CLZ equal to β.
Quadrilateral QZPB is cyclic (opposite angles are supplementary), making angle QZP equal (180 - β) which means angle KZA is β (angle in a straight line).
This means angle PZC is β (vertically opposite)
Therefore, base angles are equal and those triangles are isosceles

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Now what to do next? Maybe by introducing too many lines, I'm making things more complicated than it is.

Any help is appreciated. Thanks.

2. Hi, sqleung!

Originally Posted by sqleung
Now what to do next? Maybe by introducing too many lines, I'm making things more complicated than it is.

Any help is appreciated. Thanks.
Obviously, $\angle ZCP = \angle LCP$, so $\triangle ZPC\text{ and }\triangle LPC$ have two sides and an included angle equal, which implies that $\triangle ZPC\cong\triangle LPC$, and the remaining corresponding sides will be equal. So, $\overline{ZP} = \overline{PL}$.

Now, $\triangle ZBP\text{ and }\triangle LBP$ share a side, so you should be able to show that $\triangle ZBP\cong\triangle LBP\Rightarrow\angle LBP=\angle ZBP$

Does that help?

3. Thankyou very much for your help