Hence show that:
<BAD = 2(<CBD + <CDB)
Hello, snigdha!
Here's one solution . . . (There must be many others.)
In quadrilateral
Show that: .
Quadrilateral is a "kite".
Code:A o /|\ /θ|θ\ / | \ / | \ / | \ / | \ / θ' | θ' \ B o - - - + - - - o D * α | α * * α'|α' * * | * o C
Let
Then . where
Let
Then . where
Since is isosceles: .
. . Hence: .
And we have: .
Therefore: .
. . . . . . . . .
. . . . . .