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.