EMAT 6600
Problem Solving
From the previous lesson we can state that the sum of the area of AYCX and area of CMBN = area of ACB. To construct this into one lune, we have
Now, we ought to that the area of AIB = area of AYC + area of CNB.
We know that triangle ACB is an isosceles right triangle, as proven. So, angles CAB and CBA are congruent. Hence, subtend the same arc length. Since, AC is ½ AB as previously stated in the prior proof, 2AC is AB. AYC and CNB are subtended by angles CBA and CAB respectively. So, AIB must be subtended by angle CBA + angle CAB = angle ACB. Thus, AIB = 2AYC.¨