Two right triangles ABC and DBC are drawn on the same hypotenuse BC and on the same side of BC. If AC and BD intersect at P, prove that AP × PC = BP × PD.

Two triangles BACandBDC, right angled at Aand D respectively,are drawn on the same base BC and on the same side of BC. If AC and DB intersect at P, prove that APxPC=DPxPB.

ABC and DBC are two right angled triangles with common hypotenuse BC with their sides AC and BD intersecting at P. Prove that: AP×PC=DP×PB .

Two triangles ABC and DBC are on the same base BC and on the same side of BC in which angle A= angle D=90^@ . If CA and BD meet each other at E, show that AE. EC= BE. ED .

In the figure ABC and DBC are two right triangles. Prove that AP × PC = BP × PD

Triangles ABC and DBC are two isosceles triangles on the same base BC and their vertices A and D are on the same side of BC. If AD is extended to intersect BC at P, show that: AP bisects angle BAC and BDC.

Triangles ABC and DBC are two isosceles triangles on the same base BC and their vertices A and D are on the same side of BC. If AD is extended to intersect BC at P, show that: Delta ABP ~= Delta ACP

triangleBAC and triangleBDC are two right-triangle on the same side of the base BC.If AC ans DB intersect each other at a point P,show that APxxPC=DPxxPB .

△ABC and △DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC. If AD is extended to intersect BC at P, show that (i) △ABD≅△ACD (ii) △ABP≅△ACP (iii) AP bisects ∠A as well as △D . (iv) AP is the perpendicular bisector of BC.

In the adjacent figure, triangle ABC and triangle DBC are on the same base BC. If AD and BC intersect at O, Prove that (area of triangle ABC )/(area of triangle DBC) = (AO) /(DO)