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.`

ABC and DBC are two triangles where bar(AB) = bar(AC) and bar(DB) = bar(DC) . Show that /_ABC + /_DBC = /_ACB + /_DCB . Also S.T. AD is a bisector of /_A .

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).

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

ABC is an isoscles triangle with bar(AB)= bar(AC) and bar(AD) is one of its altitudes. Is DeltaADB ~= DeltaADC ? Why or why not?

In the adjacent figure, bar(DA) bot bar(AB), bar(CB)bot bar(AB) and AC=BD . Prove that Delta ABC ~= DeltaBAD

If ABCD is a parallelogram such that bar(AB)=bar(a), bar(BC)=bar(b)" then "bar(AC), bar(BD) are

In the adjacent figure, bar(BD) and bar(CE) are altitudes of DeltaABC such that BD= CE. Is DeltaCBD ~= DeltaBCE ? Why or why not?

In the parallelogram if bar(AB)=bar(a) and bar(AD)=bar(d) then find bar(BD) .

In the adjacent figure triangle ABC is isosceles as bar (AB) =bar(AC), bar(BA) and bar(CA) are produced to Q and P such that bar(AQ)= bar(AP) .. Show that bar(PB) =bar(QC)