Question

in triangle ABC AD and BE are altitudes prove that (ar triangleDEC )/(ar triangleABC)=(DC)^2/(AC)^2

Answer 1

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

Since ∠ADB = ∠AEB = 90°

Therefore, there must be a circle passing through point D and E having AB is 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 a 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

Now, in ΔABC and ΔDEC, we have

∠A = ∠EDC

and ∠C = ∠C (common)

therefore by AA similarity ΔABC ~ ΔDEC.

therefore arΔDEC/arΔABC = DC^2/AC^2  (ratio of areas of 2 similar triangles is equal to the ratio of the squares of corresponding sides)