In triangle ABC, AB > AC. | E is the mid-point of BC | and AD is perpendicular to BC

In triangle ABC, AB > AC. E is the mid-point of BC and AD is perpendicular to BC.
Prove that
AB² + AC² = 2.AE² + 2.BE²

Solution: In ∆ ABD AB² = AD ² + BD ²
Or AB² = AD ² + (BE + ED ) ²
Or AB² = AD ² + BE ² + 2. BE . ED + ED ²
Or AB² = AE ² + BE ² + 2. BE . ED -----(i)
In ∆ ADC AC² = AD ² + CD ²
OR AC² = AD² + ( CE - ED) ²
Or AC² = AD² + CE ² - 2. CE . ED + ED ²
Or AC² = AE² + CE ² - 2. CE . ED ------(ii)
Adding equation 1 and situation 2 we have:
AB² + AC² = AE² + AE² + BE² + CE² + 2. BE . ED - + 2. CE . ED
Or AB² + AC² = AE² + AE² + BE² + BE² + 2. BE . ED - 2. BE . ED (As BE = CE )
∴ AB² + AC² = 2 AE² + 2BE² Proved

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