Question

In triangle ABC, AD perpendicular to BC. Prove that AB² - BD2² = AC²- CD² ?
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Answer 1

Answer:

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• There is a ∆ABC with AD as the perpendicular
• AB²-BD² = AC²-CD²

$\displaystyle\underline{\bigstar\:\textsf{According to the given Question :}}$

• We shall use the Pythagoras theorem on both the triangles and then equate the equations we get from that to prove the given statement!!

In ∆ADC

$\displaystyle\sf :\implies AC^2 = AD^2+CD^2$

$\displaystyle\sf :\implies AC^2-CD^2 = AD^2\:\: -eq(1)$

In ∆ADB

$\displaystyle\sf :\implies AB^2 = AD^2+BD^2$

$\displaystyle\sf :\implies AB^2-BD^2 = AD^2\:\: -eq(2)$

$\displaystyle\underline{\bigstar\:\textsf{From 1 \& 2 :}}$

• AD² = AC² - CD²
• AD² = AB² - BD²

$\displaystyle\sf \dashrightarrow AC^2-CD^2 = AB^2 - BD^2$

$\displaystyle\sf \dashrightarrow\underline{\boxed{\sf AB^2-BD^2 = AC^2-CD^2}}$

$\displaystyle\underline{\textsf{\textbf{Hence Proved!!}}}$

Answer 2

Answer:

Hope it helps mate.....