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Question Number 87222 by TawaTawa1 last updated on 03/Apr/20

Commented by TawaTawa1 last updated on 03/Apr/20

Circles ω_1 and ω_2 intersect each other at points A and B. Point C lies on the tangent line from A to ω_1 such that ∠ABC = 90°. Arbitrary line L passes through C and cuts ω_2 at points P and Q. Lines AP and AQ cut ω_1 for the second time at points X and Z respectively. Let Y be the foot of altitude from A to L. Prove that points X, Y and Z are collinear.

$$\mathrm{Circles}\:\:\omega_{\mathrm{1}} \:\:\mathrm{and}\:\:\omega_{\mathrm{2}} \:\:\mathrm{intersect}\:\mathrm{each}\:\mathrm{other}\:\mathrm{at}\:\mathrm{points}\:\:\mathrm{A}\:\:\mathrm{and}\:\:\mathrm{B}. \\ $$$$\mathrm{Point}\:\:\mathrm{C}\:\:\mathrm{lies}\:\mathrm{on}\:\mathrm{the}\:\mathrm{tangent}\:\mathrm{line}\:\mathrm{from}\:\:\mathrm{A}\:\:\mathrm{to}\:\:\omega_{\mathrm{1}} \:\:\mathrm{such}\:\mathrm{that} \\ $$$$\angle\mathrm{ABC}\:\:=\:\:\mathrm{90}°.\:\:\mathrm{Arbitrary}\:\mathrm{line}\:\:\mathrm{L}\:\:\mathrm{passes}\:\mathrm{through}\:\:\mathrm{C}\:\:\mathrm{and} \\ $$$$\mathrm{cuts}\:\:\omega_{\mathrm{2}} \:\:\mathrm{at}\:\mathrm{points}\:\:\mathrm{P}\:\:\mathrm{and}\:\:\mathrm{Q}.\:\:\mathrm{Lines}\:\:\mathrm{AP}\:\:\mathrm{and}\:\:\mathrm{AQ}\:\:\mathrm{cut}\:\:\:\omega_{\mathrm{1}} \\ $$$$\mathrm{for}\:\mathrm{the}\:\mathrm{second}\:\mathrm{time}\:\mathrm{at}\:\mathrm{points}\:\:\mathrm{X}\:\:\mathrm{and}\:\:\mathrm{Z}\:\:\mathrm{respectively}.\:\mathrm{Let} \\ $$$$\mathrm{Y}\:\:\mathrm{be}\:\mathrm{the}\:\mathrm{foot}\:\mathrm{of}\:\mathrm{altitude}\:\mathrm{from}\:\:\mathrm{A}\:\:\mathrm{to}\:\:\mathrm{L}. \\ $$$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{points}\:\:\:\mathrm{X},\:\:\mathrm{Y}\:\:\mathrm{and}\:\:\mathrm{Z}\:\:\:\mathrm{are}\:\mathrm{collinear}. \\ $$

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