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.
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