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

triangleABC and triangleDBC are two isosceles triangles on the same base BC (see figure). Show that angelABD = angelACD .

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 angleC=90^@ then AB^2 =…………..

In Delta^s ABC and DEF, DE and AB || DE , BC=EF and BC || EF . Vertices A, B and C are joined to vertices D, E and F respectively (see figure). Show that : /_\ABC ~= /_\DEF

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

Let ABC be an isosceles triangle with BC as its base. Then, rr_(1)=

/_\ABC is an isosceles triangle with AB=AC. Draw AP _|_ BC to show that /_B=/_C

In the figure, ΔABC and ΔABD are two triangles on the same base AB. If line segment CD is bisected by bar(AB) at O, show that ar (DeltaABC) = ar (DeltaABD).

E and F are respectively the mid-points of equal sides AB and AC of DeltaABC (see figure) Show that BF = CE.