In $\Delta ABC$, $AD$ is the median to side $BC$. Prove that $AB^2 + AC^2 = 2(AD^2 + BD^2)$ (Apollonius Theorem).
Step-by-Step Solution
Key Concept: Draw altitude $AE \perp BC$. Express $AB^2$ and $AC^2$ using right triangles $ABE$ and $ACE$. Sum gives $2(AD^2 + BD^2)$.
Draw $AE \perp BC$. In right $\Delta ABE$: $AB^2 = AE^2 + BE^2 = AE^2 + (BD - ED)^2 = AE^2 + BD^2 + ED^2 - 2BD \cdot ED$. [1.0 Mark]
In right $\Delta ACE$: $AC^2 = AE^2 + EC^2 = AE^2 + (CD + ED)^2 = AE^2 + CD^2 + ED^2 + 2CD \cdot ED$. [1.0 Mark]
Add equations (note $CD = BD$): $AB^2 + AC^2 = 2(AE^2 + ED^2) + 2BD^2 = 2AD^2 + 2BD^2 = 2(AD^2 + BD^2)$. Proved! [1.0 Mark]
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🎯 Official CBSE Marking Scheme:
Drawing altitude $AE$ and setting up $AB^2$: 1.0 Mark
Setting up $AC^2$: 1.0 Mark
Adding equations to conclude $2(AD^2 + BD^2)$: 1.0 Mark
Correct Answer: