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 xx PC` = `BP xx DP`.

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.

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 .

△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 figure ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that (a r(A B C))/(a r(D B C))=(A O)/(D O) .

Delta ABC and Delta 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 E show that

DeltaABC and DeltaDBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see figure). If AD is extended to BP intersect BC at P, show that AP bisects angleA as well as angleD .

DeltaABC and DeltaDBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see figure). If AD is extended to BP intersect BC at P, show that DeltaABP~=DeltaACP

triangle ABC and triangle DBC are isosceles triangles on the same base BC. Prove that line AD bisects BC at right angles

In Figure,ABC and DBC are two isosceles triangles on the same base BC such that AB=AC and DB=CD. Prove that /_ABD=/_ACD