MathLabs

Problem 2

Let ABCABC be a triangle with AB<ACAB<AC. Let DD be the intersection point of the internal bisector of ∠BAC\angle BAC and the circumcircle of ABCABC. Let ZZ be the intersection point of the perpendicular bisector of ACAC with the external bisector of ∠BAC\angle BAC. Prove that the midpoint of segment ABAB lies on the circumcircle of triangle ADZADZ.
Step 5 of 6: Angle chase shows AD′∥BDAD'\parallel BD
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β:=∠ABC=∠ADC,∠BAD′=α+β=∠ABD  ⟹  AD′∥BD\beta:=\angle ABC=\angle ADC,\quad \angle BAD'=\alpha+\beta=\angle ABD \implies AD'\parallel BD
Detailed analysis

Write β=∠ABC=∠ADC\beta=\angle ABC=\angle ADC. Summing the angles at AA around the point (∠D′AZ+∠ZAD+∠DAB+∠BAD′=360∘\angle D'AZ+\angle ZAD+\angle DAB+\angle BAD'=360^\circ) and the internal angles of cyclic-adjacent quadrilateral AZCDAZCD (∠DCZ+∠ZAD+∠CZA+∠ADC=360∘\angle DCZ+\angle ZAD+\angle CZA+\angle ADC=360^\circ), and canceling the equal terms ∠D′AZ=∠DCZ\angle D'AZ=\angle DCZ (from the congruence) and ∠ZAD=90∘\angle ZAD=90^\circ, ∠CZA=2α\angle CZA=2\alpha, one obtains ∠BAD′=α+β\angle BAD'=\alpha+\beta, which equals ∠ABD\angle ABD. Equal alternate angles then give AD′∥BDAD'\parallel BD.