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Geometry Difficulty 4.7 AIME Prove it Greece

Let △ABΓ\triangle AB\Gamma be an acute angled triangle with circumcircle c(O,R)c(O, R). From the midpoint Δ\Delta of the side BΓB\Gamma we draw a line perpendicular to ABAB which meets ABAB at EE. If the line AOAO intersects the line ε\varepsilon at ZZ, prove that the points AA, ZZ, Δ\Delta, Γ\Gamma are cyclic.

Solution

The external angle EZ^A\mathrm{E}\hat{Z}A of the quadrilateral AZΔΓAZ\Delta\Gamma belongs to the orthogonal triangle AEZAEZ, with the acute angle EA^Z=ω\mathrm{E}\hat{A}Z = \omega equal to the angle AB^OA\hat{B}O, since OA=OBOA = OB. Hence EZ^A=90∘−ω\mathrm{E}\hat{Z}A = 90^\circ - \omega

Let the extension of the radius BOBO intersect the circle c(O,R)c(O, R) at HH. Then AΓ^H=AB^H=ωA\hat{\Gamma}H = A\hat{B}H = \omega and
90∘=BΓ^H=BΓ^A+AΓ^H⇒BΓ^A=90∘−AΓ^H=90∘−ω. 90^\circ = B\hat{\Gamma}H = B\hat{\Gamma}A + A\hat{\Gamma}H \Rightarrow B\hat{\Gamma}A = 90^\circ - A\hat{\Gamma}H = 90^\circ - \omega.
Hence EZ^A=BΓ^A\mathrm{E}\hat{Z}A = B\hat{\Gamma}A, and the quadrilateral AZΔΓAZ\Delta\Gamma is cyclic.

Figure 1
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