Prove that the bisector of the outer angle at the apex of an isosceles triangle is parallel to its base.

Solution.

1. Consider the triangle ABC. <B = 180 ° – <A – <C = 180 ° – 2 <C

2. <ABP = 180 ° – <B = 180 ° – 180 ° +2 <C = 2 <C.

3. By the condition of the problem <ABK = 1/2 <ABP = <C = <B.

4. Consider <C and <CBK

<CBK = <ABK + <B

<ABK = <C (from (3)), then <CBK = <C + <B

5. <CBK + <ACB = <C + <B + <C, where <C = <A, that is

<CBK + <ACB = <C + <B + <A = 180 ° as the sum of the angles of the triangle.

In this case, <CBK and <ACB are internal one-sided angles, their sum is 180 °, which means that the straight lines BK and AC are parallel, which was required to be proved.