Let ABCD is a Cyclic Quardrilateral. If Angle A = 80 degree?

Question

and AC Biscects Angle BAD and Angle BCD. How to Prove that AC is the diameter.

Answer 1

as angle A=80 degree
angle C=100 degree

now AC Biscects Angle BAD and Angle BCD

so in the triangle ACD

angle c is 90 degree draw it to get better understanding

we knw that angle in a semicircle is 90 degree only if it is subtended by diameter

done!!!!!!

Answer 2

Let

α = measure of angle A
γ = measure of angle C

Given

Quadrilateral ABCD is cyclic
AC bisects angles A and C
α = 80°

Prove AC is a diameter.

We have

AC bisects angles A and C. Therefore:


Every triangle has 180°. So:


But since ABCD is a cyclic quadrilateral
2*
Angles B and D subtend 2*90° = 180° of arc. So they subtend a diameter.

Angles B and D subtend AC. Therefore AC is a diameter.

Answer 3

If AC bisects both BAD and BCD, then triangles ACD and ACB are congruent by ASA. Since the quadrilateral is symmetric about AC, AC must be a diameter. A secondary argument is that AC is the perpendicular bisector of BD, and must therefore pass through the center of the circle. Any line crossing a circle through its center is a diameter.