In the triangle ABC,∠BAC is acute, the angle bisector of ∠BAC meets BC at D,K is the foot of the perpendicular from B to AC, and ∠ADB=45o. Point P lies between K and C such that ∠KDP=30o. Point Q lies on the ray DP such that DQ=DK. The perpendicular at P to AC meets KD at L. Prove that PL2=DQ⋅PQ. geometryanglesAngle Chasingangle bisectorequal segments