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I'm having trouble solving the following geometry problem and would appreciate any help. I ended up proving some other two lines were parallel instead of the desired ones. Please feel free to change the title to a more descriptive one, as I didn't know how to word such a problem. Thank you.

Let $P$ be the point on the circumcircle of $\triangle{ABC}$ such that the perpendicular from this point to $\overleftrightarrow{BC}$ is also tangent to the circle. Draw the perpendicular to $\overleftrightarrow{AB}$ through $P$ and label the intersection point $Z$ as shown. Prove that $\overleftrightarrow{AP}\parallel\overleftrightarrow{ZX}$. enter image description here

Attempt:

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Alex D
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2 Answers2

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Simply note that quadrilateral $XBZP$ is cyclic as $\angle BZP + \angle BXP = 180^\circ$. Then $\angle BZX = \angle BPX$, and $\angle BPX = \angle BAP$ by the Inscribed Angle Theorem(the limiting case).

Thus, as $\angle BZX = \angle BAP$, $$\overline{AP} \parallel \overline{ZX}$$.

  • Thank you SO much. We're covering simpson lines and I thought the solution had something do to with them but this is so simple. +1 for me – Alex D Apr 30 '18 at 22:30
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Another proof. Note again that quadrilateral $XBZP$ is cyclic. Therefore $\angle ZXB = \angle ZPB$. Now draw line $\overline{PO}$ and let $D = \overline{AB} \cap \overline{PO}$. The circumcenter and orthocenter are isogonal conjugates, therefore $\angle APD = \angle ZPB$. Including the fact that $\overline{PD} \parallel \overline{XB}$, it follows that $$\overline{AP} \parallel \overline{ZX}$$ .