The perpendicular on the base of a ∆ intersects at so that = 3. Prove that 2()2 = 2()2 + 2 .

We are given that D is perpendicular to and = 3.Here, = + ⇒ = 3 + (given)⇒ = 4⇒ = 1/4 So, = 3⇒ = 3/4 So, ∆ and ∆ are right-angled triangles.We can apply Pythagoras’ theorem to these triangles.∆, 2 + 2 = 2 … eq(1)⇒ + ( 3/4BC)2 = AB2= AD2 + 9/16 BC2 + AB2⇒ 2 = AB2 - 9/16BC2∆, 2 + 2 = AC22 + (1/4BC)2 = AC22 + 1/16BC2 = AC22 − 9/16B2 +1/ 16 2 = AC2⇒ 2 − 2 = 9/16 2 - 1/16BC2⇒ 2 − 2 = 8/162⇒ 2(2 − 2 ) = 2⇒ 22 − 22 = 2⇒ 2()2 = 2()2 + 2Hence, proved.

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The perpendicular 𝐴𝐷 on the base 𝐵𝐶 of a ∆𝐴𝐵𝐶 intersects 𝐵𝐶 at 𝐷 so that
𝐷𝐵 = 3𝐶𝐷. Prove that 2(𝐴𝐵)² = 2(𝐴𝐶)² + 𝐵𝐶² .

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Detailed Solution

We are given that 𝐴D is perpendicular to 𝐵𝐶 and 𝐷𝐵 = 3𝐶𝐷.

Here, 𝐵𝐶 = 𝐵𝐷 + 𝐶𝐷

⇒ 𝐵𝐶 = 3𝐶𝐷 + 𝐶𝐷 (given)

⇒ 𝐵𝐶 = 4𝐶𝐷

⇒ 𝐶𝐷 = 1/4 𝐵𝐶

So, 𝐷𝐵 = 3𝐶𝐷
⇒ 𝐷𝐵 = 3/4 𝐵𝐶
So, ∆𝐴𝐵𝐷 and ∆𝐴𝐷𝐶 are right-angled triangles.
We can apply Pythagoras’ theorem to these triangles.
∆𝐴𝐵𝐷, 𝐴𝐷² + 𝐵𝐷² = 𝐴𝐵²… eq(1)

⇒ 𝐴𝐷 + ( 3/4BC)²= AB²
= AD² + 9/16 BC² + AB²
⇒ 𝐴𝐷² = AB² - 9/16BC²

∆𝐴𝐶𝐷, 𝐴𝐷² + 𝐶𝐷² = AC²

𝐴𝐷² + (1/4BC)² = AC²

𝐴𝐷² + 1/16BC² = AC²

𝐴𝐵² − 9/16B𝐶² +1/ 16 𝐵𝐶² = AC²
⇒ 𝐴𝐵² − 𝐴𝐶² = 9/16 𝐵𝐶²- 1/16BC²
⇒ 𝐴𝐵² − 𝐴𝐶² = 8/16𝐵𝐶²
⇒ 2(𝐴𝐵² − 𝐴𝐶² ) = 𝐵𝐶²
⇒ 2𝐴𝐵² − 2𝐴𝐶² = 𝐵𝐶²
⇒ 2(𝐴𝐵)² = 2(𝐴𝐶)² + 𝐵𝐶²
Hence, proved.