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

A B C\ a n d\ D B C are two isosceles triangles on the same bas B C and vertices A\ a n d\ D are on the same side of B C . If A D is extended to intersect B C at P , show that A P bisects " "/_A as well as /_D and A P is the perpendicular bisector of B C

DeltaA B C and DeltaD B C are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see Fig. 7.39). If AD is extended to intersect BC at P, show that (i) \ DeltaA B D~=DeltaA C D (ii) DeltaA B P~=DeltaACP (iii) AP bisects ∠ A as well as ∠ D (iv) AP is the perpendicular bisector of BC

A B C\ a n d\ D B C are two isosceles triangles on the same bas B C and vertices A\ a n d\ D are on the same side of B C . If A D is extended to intersect B C at P , show that A B D\ ~= A C D (ii) A B P\ ~= A C P

ABC and DBC are two isosceles triangles on the same base BC (see Fig. 7.33). Show that /_A B D\ =/_A C D

If D is a Point on the side BC of a Delta ABC such that AD bisects angleBAC . Then

ABC and DBC are two triangle on the same base BC such that A and D lie on the opposite sides of BC,AB=AC and DB =DC ,Show that AD is the perpedicular bisector of BC.

Delta ABC and Delta DBC are on same base BC and their vertices A and D are on opposite sides of BC. It is given that: area of Delta ABC = " area of " Delta DBC Prove that BC bisects the line segment AD.

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) .

triangleABC and triangleDBC are two triangles on the same base BC. A and D lies on opposite sides of BC. Prove that (ar(triangleABC))/(ar(triangleDBC))= (AO)/(DO)