Question

In triangle ABC, AD is drawn
perpendicular to Bc. If
BD :CD = 3:1, then prove
that Bc²= 2 (AB²-AC²)​

Answer 1

Answer:

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Answer 2

Explanation:

Given:A triangle ABC,AD perpendicular to BC and BD:DC=3:1

To prove :2(AB^2-AC^2)=BC^2

Since: BD:DC=3:1

So,BD=3x and DC=x

In right triangle ADB

AB^2=BD^2+AD^2

=>AB^2=9x^2+AD^2......(1)

In right triangle ADC

AC^2=DC^2+AD^2

=>AC^2=X^2+AD^2.......(2)

Eq(1) -Eq(2)

AB^2-AC^2=9x^2-x^2+AD^2-AD^2

=>2(AB^2-AC^2)=>2×8x^2

2(AB^2-AC^2=16x^2=(4x^2)=BC^2

Hence proved

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