Question

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

Answer 1

Step-by-step explanation:

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Answer 2

Step-by-step explanation:

triangle ABC

AB>AC                   E mid-point of BC

AD ⊥ BC

In ΔABD ,by Pythagoras Theorem:  

AB²=AD²+BD²

     =AD²+(BE+ED)²                              {BD=BE+ED}

AB²=AD²+BE²+ED²+2BE.ED        {1}

In ΔADC ,by Pythagoras Theorem:  

AC²=AD²+DC²

     =AD²+(CE-ED)²                {DC=CE-ED}

AC²=AD²+CE²+ED²-2CE.ED      {2}

Add {1} and {2}:

AB²+AC²=2AD²+BE²+CE²+2ED²-2CE.ED+2BE.ED

AB²+AC²=2AD²+BE²+BE²+2ED²-2BE.ED+2BE.ED      {BE=CE}

AB²+AC²=2(AD²+ED²) +2BE²

AB²+AC²=2AE² +2BE²

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