1.
Draw a circle with center C And radius CA, as shown in the attached picture.
Let the lengths of radii AO, OB, OC be R. Triangle ABC is inscribed in the circle with center O and one of its sides is a diameter, this means that the angle ACB is a right angle.
|AO|=|OC|=R, by the Pythagorean theorem |AC|=[tex] \sqrt{2}R [/tex].
these are all shown in the picture.
2.
Area of triangle ABC is 1/2 * 2R * R= R^2
3.
Let the area between arc BXA and chord AB be Y. (the yellow region).
and let G be the shaded region between arcs AB and AXB.
G=1/2(Area circle with center O)-Y
=[tex] \frac{1}{2} \pi R^{2}-Y [/tex]
To find Y:
Notice that the area of the sector ACB is 1/4 of the area of circle with center C, since m(ACB) is 1/4 of 360 degrees.
So Area of sector ACB = [tex]\frac{1}{4} \pi (\sqrt{2} R)^{2}=\frac{1}{4} \pi*2 R^{2} =\frac{1}{2} \pi R^{2}[/tex]
Y =area of sector ABC-Area(triangle ABC)=[tex]\frac{1}{2} \pi R^{2}- \frac{1}{2}*2R*R=\frac{1}{2} \pi R^{2}- R^{2} [/tex]
4.
Finally,
[tex]G=\frac{1}{2} \pi R^{2}-Y=\frac{1}{2} \pi R^{2}-(\frac{1}{2} \pi R^{2}- R^{2})=R^{2}[/tex]
This proves that the 2 shaded regions have equal area.
