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Angle Sum Property

In triangle ...

Question

In triangle $A B C, ∠ B = 90^{0}$ and $D$ is mid-point of $B C$ . Prove that: $A C^{2} = A D^{2} + 3 C D^{2}$

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Solution

In $Δ A B C,$
${A C}^{2} = {A B}^{2} + {B C}^{2} = {A B}^{2} + {4 C D}^{2} . . . . [B C = 2 C D]$

In $Δ A B D,$
${A D}^{2} = {A B}^{2} + {B D}^{2} = {A B}^{2} + {C D}^{2}$

Subtracting the above equations, we get
${A C}^{2} - {A D}^{2} = {A B}^{2} + {4 C D}^{2} - ({A B}^{2} + {C D}^{2}) {A C}^{2} - {A D}^{2} = {3 C D}^{2} {A C}^{2} = {A D}^{2} + {3 C D}^{2}$

Hence proved.

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