Given, the perpendicular from A on side BC of a Δ ABC intersects BC at D such that;

DB = 3CD.

In Δ ABC,

AD ⊥BC and BD = 3CD

In right angle triangle, ADB and ADC, by Pythagoras theorem,

AB^{2} =^{ }AD^{2} + BD^{2}……………….**(i)**

AC^{2} =^{ }AD^{2} + DC^{2} …………..**(ii)**

Subtracting equation **(ii)** from equation **(i)**, we get

AB^{2} – AC^{2} = BD^{2} – DC^{2}

= 9CD^{2} – CD^{2} [Since, BD = 3CD]

= 8CD^{2}

= 8(\( \dfrac{BC}{4}\))^{2 }[Since, BC = DB + CD = 3CD + CD = 4CD]

Therefore, AB^{2} – AC^{2} = BC^{2}/2

⇒ 2(AB^{2} – AC^{2}) = BC^{2}

⇒ 2AB^{2} – 2AC^{2} = BC^{2}

2AB^{2} = 2AC^{2} + BC^{2}.

In an trapezium ABCD, it is given that AB ∥ CD and AB = 2CD. Its diagonals AC and BD intersect at the point O such that ar(∆AOB) = 84 cm^{2}. Find ar(∆COD)

Find the length of the altitude of an equilateral triangle of side 2a cm.

A ladder 10 m long reaches the window of a house 8 m above the ground. Find the distance of the foot of the ladder from the base of the wall.

In the given figure, DE ∥ BC such that AD = x cm, DB = (3x + 4) cm, AE = (x + 3) cm and EC = (3x + 19) cm. Find the value of x.