`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.

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

In the adjoning figure, D and E are the points on the sides AB and AC respectively of Delta ABC and area of Delta BCE = " area of " Delta BCD . Prove that DE ||BC

In the adjoining figure, triangle ABC and triangleDBC are on the same base BC with A and D on opposite sides of BC such that ar(triangleABC)=ar(triangleDBC) . Show that BC bisect AD.

D is a point on the base BC of Delta ABC.AD is produced upto E such that DE = AD. Prove that: area of Delta BCE = " area of " Delta ABC

D and E are points on sides AB and AC respectively of Delta ABC such that ar( Delta DBC) = ar( Delta EBC). Prove that DE I I BC.

Delta ABC is an equilateral triangle and D and E are midpoints of AB and BC respectively. Then the area of Delta ABC : the area of the trapezium ADEC is

In the same figure, Delta ABC and Delta DBC are on the same base BC . If AD is intersects BC at O, prove that (ar(Delta ABC))/(ar (Delta DBC))=(AO)/(DO)

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 the given figure,Delta ABC and Delta DBC are on the same base BC.If AD intersects BC at O , prove that (ar(Delta ABC))/(ar(Delta DBC))=(AO)/(DO)

△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.