Question

In a cyclic quadrilateral ABCD, the diagonal AC bisects the $\angle BCD$. Prove that the diagonal BD is parallel to the tangent to the circle at point A.

Answer 1

Given, ∠ACD=∠ACB.......(1)Here we can see that $\angle NAD=\angle ABD$[Angle in the alternate segment]

Similarly , $\angle MAB=\angle ADB$

Also AB is common arc and $\angle ADBand\angle ACB$ are the angle in the same segment

so ∠ADB=∠ACB

Similarly ∠ABD=∠ACD

But using equation 1 we can say that $\angle 1=\angle 2$

So ∠NAD=∠ADB
[Alternate interior angle]

So $MN\parallel BD$