A line segment BK is an angle bisector of ΔABC. A line KM intersects side BC such, that BM = MK. Prove: KM ∥ AB.

Answer:∠BKM= ∠ABK Therefore AB ║KM (∵ ∠BKM= ∠ABK  and lies between AB and KM and BK is the transversal line)m∠MBK ≅ m∠BKM (Angles opposite to equal side of ΔBMK are equal)Step-by-step explanation:Given: BK is an angle bisector of Δ ABC. and line KM intersect BC such that, BM = MKTO prove: KM ║ABNow, As given in figure 1, In Δ ABC, ∠ABK = ∠KBC (∵ BK is angle bisector) Now in Δ BMK, ∠MBK = ∠BKM (∵ BM = MK and angles opposite to equal sides of a triangle are equal.)Now ∵ ∠MBK = ∠BKM and  ∠ABK = ∠KBM ∴ ∠BKM= ∠ABK Therefore AB ║KM (∵ ∠BKM= ∠ABK and BK is the transversal line) Hence proved.