Question

In ∆ABC, AD and BE are altitude. Prove that\[\frac{ar\left( ∆ DEC \right)}{ar\left( ∆ ABC \right)} = \frac{{DC}^2}{{AC}^2}\]

Answer 1

Δ

Given: ΔABC in which AD and BE are altitudes on sides BC and AC respectively.

Since ∠ADB = ∠AEB = 90°, there must be a circle passing through point D and E having AB as diameter.

We also know that, angle in a semi-circle is a right angle.

Now, join DE.

So, ABDE is a cyclic quadrilateral with AB being the diameter of the circle.

∠A + ∠BDE = 180° [Opposite angles in a cyclic quadrilateral are supplementary]

⇒ ∠A + (∠BDA + ∠ADE) = 180°

⇒ ∠BDA + ∠ADE = 180° − ∠A ..... (1)

Again,

∠BDA + ∠ADC = 180° [Linear pair]

⇒ ∠BDA + ∠ADE + ∠EDC = 180°

⇒ ∠BDA + ∠ADE = 180° − ∠EDC ..... (2)

Equating (1) and (2), we get

180° − ∠A = 180° − ∠EDC

⇒ ∠A = ∠EDC

Similarly, ∠B = ∠CED

Now, in ΔABC and ΔDEC, we have

∠A = ∠EDC

∠B = ∠CED

∠C = ∠C

∴ ΔABC ∼ ΔDEC

`⇒ \text{(Area of Δ DEC )}/ \text{(Area of Δ ABC )}= ((DC)/(AC))^2`