`DeltaABC and DeltaDBC` are two isosceles triangles on thesame base BC (see figure). Show that `/_ ABD = /_ ACD.`

ABC and DBC are two isosceles triangles on the same base BC. Show that ABD = ACD.

ABC and DBC are two isosceles triangles on the same base BC. Show that, triangleABD = angleACD

triangleABC and triangleDBC 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) triangleABD ~= triangleACD (ii) triangleABP ~= triangleACP (iii) AP bisects angleA as well as angleD (iv) AP is the perpendicular bisector of BC.

In the adjacent figure DeltaABC and DeltaDBC are two triangles such that bar(AB) = bar(BD) and bar(AC) = bar(CD) . Show that DeltaABC ~= DeltaDBC.

DeltaABC is an isosceles triangle in which AB = AC. Show that /_ B = /_ C.

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

In the figure ABC and DBC are two triangles on the same base BC. If AD intersects BC at 0, show that ("Area"(DeltaABC))/("Area"(DeltaDBC))=(AO) /(DO) .

In Fig , ABC and DBC are two triangles on the same base BC. If AD intersects BC,at O , show that (ar(ABC))/(ar(DBC))=(AO)/(DO)

D is a point on side BC ΔABC such that AD = AC (see figure). Show that AB > AD.

In the given figure, ABC and ABD are two triangles on the same base AB . If line - segment CD is bisected by AB at O, show that ar (ABC) = ar (ABD)