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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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