In `Delta ABC` bisectors of `/_B and /_C` meet each other at X. Line AX cuts side BC at Y. Hence `(AX)/(XY)` = ?

Bisector of /_B and /_C in DeltaABC meet each other at P. Line AP cuts the sides BC at Q. Prove: (AP)/(PQ)=(AB+AC)/(BC) .

In the figure bisectors of /_B and /_C of DeltaABC intersect each other in point X. Line AX intersects side BC in pont Y. AB=5, AC=4, BC=6 then find (AX)/(XY)

In the figure bisectors of /_B and /_C of DeltaABC intersect each other in point X. Line AX intersects side BC in pont Y. AB=5, AC=4, BC=6 then find (AX)/(XY)

In the adjoining figure bisectors of angleBand angleC intersect each other in point X. Line AX intersects side BC in point Y. AB=5, AC=4, BC=6 "then find" (AX)/(XY) .

Bisector of angleB" and "angle C " in "DeltaABC meet each other at P. Line Ap cuts the side BC at Q. Then prove that : (AP)/(PQ)=(AB+BC)/(BC)

In a Delta ABC the bisector of angles B and C lie along the lines x=y and y=0. If A is (1,2), then sqrt(10)d(A,BC) where d (A, BC)represents distance of point A from side BC

In a Delta ABC the bisector of angles B and C lie along the lines x = y and y = 0 . If A is (1, 2) , then sqrt10d(A,BC) where d (A, BC)represents distance of point A from side BC

A B is a line segment, A X and B Y are two equal line segments drawn on opposite sides of line A B such that AX||BY . If A B and X Y intersect each other at P , prove that triangle A P X~= triangle B P Y . A B and X Y bisect each other.

The internal bisector of angle A of Delta ABC meets BC at D and the external bisector of angle A meets BC produced at E. Prove that (BD)/(BE) = (CD)/(CE) .