Question

in triangle abc ad and be are altitudes. prove that ar(triangle abc)/ar(dec)=ac^2/dc^2

Answer 1

Given, a 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

Similarly, ∠B = ∠CED

Now, in ΔABC and ΔDEC, we have

∠A = ∠EDC

∠B = ∠CED

and ∠C = ∠C

Therefore, ΔABC is similar to ΔDEC by AAA

Hence proved.

Answer 2

Answer:Given, a 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

Similarly, ∠B = ∠CED

Now, in ΔABC and ΔDEC, we have

∠A = ∠EDC

∠B = ∠CED

and ∠C = ∠C

Therefore, ΔABC is similar to ΔDEC by AAA

Hence proved.

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