Question

In right-angled triangle ABC in which $\angle C={90}^o$, if D is the mid-point of BC, prove that $A{B}^2=4A{D}^2-3A{C}^2$.

Answer 1

To prove:- $A{B}^2=4A{D}^2-3A{C}^2$

Proof:-

In $△ACD$

$A{C}^2+C{D}^2=A{D}^2$ (Pythagoras Theorem)\

$A{C}^2+(\frac{BC}{2}{)}^2=A{D}^2$ (Pythagoras Theorem)-------------(1)

In $△ABC$,

$A{C}^2+B{C}^2=A{B}^2$

$\Rightarrow B{C}^2=A{B}^2-A{C}^2$------------(2)

Substituting Equation(2) in (1), we get

$\Rightarrow A{C}^2+\frac{A{B}^2}{4}-\frac{A{C}^2}{4}=A{D}^2$

$\Rightarrow \frac{3A{C}^2}{4}+\frac{A{B}^2}{4}=A{D}^2$

$\Rightarrow A{B}^2=4A{D}^2-3A{C}^2$