Question

In ∆ ABC, seg AD perpendicular to BC. DB=3CD. Prove that 9(AB square-AC square)=8BD square.​

Answer 1

Given : ∆ ABC, seg AD perpendicular to BC. DB=3CD

To find : Prove that 9(AB² - AC²) = 8BD²

Solution:

Applying Pythagoras theorem in right angle Δ ADB

AD² = AB²  - DB²

DB = 3CD

=> AD² = AB²  - (3CD)²

=> AD² = AB²  - 9CD²

AD² = AC² - CD²  in right angle Δ ADC

Equating AD²

AB²  - 9CD² =  AC² - CD²

=>AB² - AC² = 8CD²

CD  =BD/3

=> AB² - AC² = 8(BD/3)²

=> AB² - AC² = 8BD²/9

=> 9(AB² - AC²) = 8BD²

QED

Hence proved

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