# How will you prove the below identiy ?

**Using inscribed angle theorem we get:-
**

2B = sum of measure of arcs (AE+ED+DC)

2E = sum of measure of arcs (AB+BC+CD)

2C = sum of measure of arcs(BA+AE+ED)

since sum of measure of all arcs in a circle is 360 degrees.

=> (B+E) = (2CD+AB+BC+AE+ED) = 0.5(360+ CD)

=> (C+E) = (2AB+AE+ED+BC+CD) = 0.5(360 + AB)

Taking sin of both sides

sin(B+E) = sin (180+CD/2) = -sin(CD/2)

sin(C+E) = sin (180+AB/2) = -sin(AB/2)

sin(B+E) -sin(CD/2)

=> ________ = ___________ ——————————-(1)

sin(C+E) -sin(AB/2)

By applying cosine rule on the constructed triangles OCD an OAB, we get:-

a^2 = 2*r*r – 2*r*r cos (measure of arc CD) = 2*r*r(1 – cos (measure arc CD)) ———— (2)

d^2 = 2*r*r – 2*r*r cos (measure of arc AB) = 2*r*r(1 – cos (measure arc AB)) —————-(3)

Dividing equation (2) by (3) and taking square root of both sides **(considering negative value of the square root only to substitute in equation 1)**

a sqrt(1 – cos CD)

__ = ___________________

d sqrt(1 – cos AB)

sqrt(1 – (1 – 2*sin^2 CD/2))

= _____________________________________

sqrt(1 – (1 – 2*sin^2 CD/2))

sqrt(2) * -sin(CD/2)

= ______________________ ——————————-(4)

sqrt(2) * -sin(AB/2)

FROM EQUATIONS 1 AND 4 we get:-

**sin(B+E) a**

**________ = ______**

**sin(C+E) d**