Special Case of Morley's Theorem The Law of Sines for ∆AQC in the figure yields AQ/sin...

Special Case of Morley's Theorem The Law of Sines for ∆AQC in the figure yields AQ/sin 30° = AC/sin 135°. From this, derive the value AQ =. If M is the midpoint of side AB, show that AM = = AQ. Thus, ∆ARQ is congruent to ∆ARM. and ∠RQA is a right angle, (Complete these details.) Compute the resulting angle measures of ∆PQR to show it is equiangular, hence, equilateral.