Question

In a right triangle ABC right angled at A, D is any point on AB. Prove that DC^2 = BC^2 + BD^2 - 2AB ×BD

Answer 1

In right ∆ ABC we get,

→ BC² = AB² + AC² [Pythagoras Theorem]

→ 0 = BC² - AB² - AC² __(1)

In right ∆ ACD we have,

→ CD² = AC² + AD² [Pythagoras Theorem] __(2)

On adding both equations we get,

→ CD² = AC² + AD² + BC² - AC² - AB²

→ CD² = BC² + AD² - AB²

→ CD² = BC² + (AB - BD)² - AB²

→ CD² = BC² + AB² + BD² - 2 · AB · BD - AB²

→ CD² = BC² + BD² - 2·AB·BD

Q.E.D

Answer 2

GIVEN: A right ∆ABC, right angled at A; having a point D on AB.

Prove: DC² = BC² + BD² - 2AB × BD

Proof:

In ∆ABC

By Pythagoras theoram

→ BC² = AB² + AC² ________ (eq 1)

In ∆ACD

By Pythagoras theoram

→ DC² = AC² + AD² ________ (eq 2)

Sub (eq 1) from (eq 2)

→ DC² - BC² = AC² + AD² - AB² - AC²

→ DC² - BC² = AD² - AB²

Now..

AD = AB - BD

→ DC² - BC² = (AB - BD)² - AB²

→ DC² - BC² = AB² + BD² - 2AB × BD - AB²

→ DC² - BC² = BD² - 2AB × BD

→ DC² = BC² + BD² - 2AB × BD

__________ [ PROVED ]

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