Question

Grade 10
Cbse
Chapter Traingles

The perpendicular from A on side BC of a ∆ABC intersect BC at D such that BD = 1/3 BC. Prove that 3AB² = 3AC² - BC²

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Answer 1

Hello akshi
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Your answer :-)

Check attachment for the figure.

Given that : -

In ∆ABC,

AD ⊥ BC

BD = ⅓BC

To prove : - 3AB² = 3AC² - BC²

Now,

BC = BD + CD

=> BC = ⅓ BC + CD

=> CD = BC - ⅓BC

=> CD = ⅔ BC

Now,

Using PYTHAGORAS THEOREM,

AB² = AD² + BD²

=> AB² = AC² - CD² + BD² [In ∆ACD, AD² = AC² - CD²]

=> AB² = AC² - (CD² - BD²)

=> AB² = AC² - (CD + BD)(CD - BD)

=> AB² = AC² - BC × (⅔ BC - ⅓ BC). [BD = ⅓ BC & CD = ⅔ BC, and (CD + BD) = BC]

=> AB² = AC² - BC × BC/3

=> AB² = AC² - BC²/3

Multiplying throughout by 3 , we get ,

3AB² = 3AC² - BC²

HOPE IT HELPS ;-)

Answer 2

HEYA!
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❇TRIANGLES❇
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⭐CONSULT ATTACHMENT FOR THE SOLUTION⭐

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