[[!redirects quarter circle solution]] # Solution to the Quarter Circle Overlapping a Circle Problem ## Using Angles in a Circle +-- { .image } [[AnglesInCircleQCircleSolution.png:pic]] =-- Using the labels from the above diagram, the total angle that we want to calculate can be written as: $$ A \hat { E } D - A \hat { E } C + A \hat { E } B $$ Now, angles $A \hat{B } E$ and $A \hat{D } E$ are the same because they are [[angles in the same segment|subtended at the orange circle from the same chord]]. Then angles $A \hat{D } E$ and $A \hat{E } D$ are the same because triangle $ A D E $ is isosceles as sides $ A D $ and $ A E $ are both radii of the black quarter circle. Since triangle $A B E $ is a right angled triangle, $A \hat { E } B + A \hat { E } D = 90^\circ$. Since triangle $A C E $ is isosceles, angle $ A \hat{ E } C = 45^\circ$. Combining these, we get: $$ A \hat { E } D - A \hat { E } C + A \hat { E } B = A \hat { B } E + A \hat { E } B - 45^\circ = 90^\circ -45^\circ = 45^\circ $$ ## Using the [[Invariance Principle]] In this problem, the size of the orange circle can change so we can invoke the [[Agg invariance principle]]. As it varies, the location of point $B$ (in the above diagram) slides between $A$ and $C$. By sliding it to $C$, the angle $ A \hat{ E } B$ becomes $45^\circ$ while the other two angles become zero. +-- { .image } [[AnglesInCircleQCircleSolutionII.png:pic]] =--