In `DeltaABC`, AD is drawn perpendicular to BC. If `BD:CD=3:1`, then prove that `BC^(2)=2(AB^(2)-AC^(2))`.

In an equilateral triangle ABC, AD is drawn perpendicular to BC meeting BC in A Prove that AD^(2) = x BD^(2) Find x.

In the figure, If AC=BD then prove that AB=CD.

In Delta ABC , BD : CD = 3 : 1 and AD _|_ BC . Prove that 2(AB^(2) - AC^(2)) = BC^(2) .

In the given fig. if AD bot BC Prove that AB^(2) + CD^(2) = BD^(2) + AC^(2) .

ABCD is a rhombus. Prove that AC^(2) + BD^(2) = 4 AB^(2)

The perpendicular from A on side BC of a DeltaABC intersects BC at D such the DB = 3 CD . Prove that 2 AB^(2)=2AC^(2)+BC^(2)

In an equilateral triangle ABC, AD _|_ BC . Prove that : AB^(2) + CD^(2) = 5/4 AC^(2)

In DeltaABC , AD is the perpendicular bisector of BC (See adjacent figure). Show that DeltaABC is an isosceles triangle in which AB = AC.

In a DeltaABC , if D is the middle point of BC and AD is perpendicular to AC, then cos C equals

In Delta ABC , AD bot BC and AD^(2)= BD xx CD . Prove that AB^(2) + AC^(2) = (BD + CD)^(2)