MathLabs

Problem 3

Let ABAB and ACAC be two distinct rays not lying on the same line, and let ω\omega be a circle with center OO that is tangent to ray ACAC at EE and ray ABAB at FF. Let RR be a point on segment EFEF. The line through OO parallel to EFEF intersects line ABAB at PP. Let NN be the intersection of lines PRPR and ACAC, and let MM be the intersection of line ABAB and the line through RR parallel to ACAC. Prove that line MNMN is tangent to ω\omega.
Step 3 of 5: Compute the half-angle at OO
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∠M′OP=∠M′OF+∠FOP=90∘−12∠XOE,∠AYM′=12∠XOE=∠NOE\angle M'OP=\angle M'OF+\angle FOP=90^\circ-\tfrac12\angle XOE,\quad \angle AYM'=\tfrac12\angle XOE=\angle NOE
Detailed analysis

Since 2∠M′OP=∠FOX+∠FOP+∠EOQ=180∘−∠XOE2\angle M'OP=\angle FOX+\angle FOP+\angle EOQ=180^\circ-\angle XOE, we get ∠M′OP=90∘−12∠XOE\angle M'OP=90^\circ-\tfrac12\angle XOE. Because ∠AYM′\angle AYM' and ∠M′OP\angle M'OP are complementary, ∠AYM′=12∠XOE=∠NOE\angle AYM'=\tfrac12\angle XOE=\angle NOE, so AA and NN are corresponding points under the similarity of △M′YP\triangle M'YP and △EOQ\triangle EOQ.