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

Prove that

AB^{2} + AC^{2} = 2.AE^{2} + 2.BE^{2}

Solution:In ∆ ABD AB^{2} = AD ^{2} + BD ^{2}

Or AB^{2} = AD ^{2} + (BE + ED ) ^{2}

Or AB^{2} = AD ^{2} + BE ^{2} + 2. BE . ED + ED ^{2}

Or AB^{2} = AE ^{2} + BE ^{2} + 2. BE . ED —–(i)

In ∆ ADC AC^{2} = AD ^{2} + CD ^{2}

OR AC^{2} = AD^{2} + ( CE – ED) ^{2}

Or AC^{2} = AD^{2} + CE ^{2} – 2. CE . ED + ED ^{2}

Or AC^{2} = AE^{2} + CE ^{2} – 2. CE . ED ——(ii)

Adding equation 1 and situation 2 we have:

AB^{2} + AC^{2} = AE^{2} + AE^{2} + BE^{2} + CE^{2} + 2. BE . ED – + 2. CE . ED

Or AB^{2} + AC^{2} = AE^{2} + AE^{2} + BE^{2} + BE^{2} + 2. BE . ED – 2. BE . ED (As BE = CE )

∴ AB^{2} + AC^{2} = 2 AE^{2} + 2BE^{2} __Proved__

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