The bisector of the outer angle at the vertex C of the triangle ABC intersects the circumcircle

| August 10, 2021 | education

The bisector of the outer angle at the vertex C of the triangle ABC intersects the circumcircle at point D. Prove that AD = BD.

Let the angle ECD = ACD = X0.

Since CD is the bisector of the ACE angle, the ACE angle = 2 * X0.

The ACB angle is adjacent to the ACE angle, the sum of which is 180, then the ACB angle = (180 – 2 * X).

The inscribed angle ACB rests on the arc AB as well as the inscribed angle ADB, then the angle ADB = ACB = (180 – 2 * X).

The inscribed angle ACD = X0 and rests on the arc AD as well as the inscribed angle ABD, then the angle ABD = ACD = X0.

Then in the triangle ABD the angle BAD = (180 – ADB – ABD) = (180 – (180 – 2 * X) – X) = X0.

Since the angle BAD = ABD = X0, then the triangle ABD is isosceles, AD = BD, which was required to be proved.



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