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 `

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.

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

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

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.

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) .

The given figure shows two isosceles triangles ABC and DBC with common base BC. AD is extended to intersect BC at point P. Show that: Delta ABD = Delta ACD (ii) Delta ABP = Delta ACP (iii) AP bisects angle BDC (iv) AP is perpendicular 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.

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)

Construct two isosceles triangles ABC and DBC, with common base BC and the points A and D on opposite sides of BC, given that BC = 8 cm, AB = AC = 4.5 cm, and DB = DC = 5 cm.

In isosceles triangle ABC, D is a point on the base BC produced. Prove that AD gt AB .