Question

Prove $2A{B}^2=2A{C}^2+B{C}^2$ in the following figure.

Answer 1

BD = 3 CD
Let CD = x
$\Rightarrow BD=3x$ and $BC=4x$
$\therefore CD=\frac{1}{4}BC\dots \dots \left(i\right)BD=\frac{3}{4}BC\dots \dots \left(ii\right)Inright\Delta ADB,A{D}^2=A{B}^2-B{D}^2\dots \dots \left(iii\right)Inright\Delta ADC,A{D}^2=A{C}^2-C{D}^2\dots \dots \left(iv\right)From\left(iii\right)and\left(iv\right),wegetA{B}^2-B{D}^2=A{C}^2-C{D}^2\therefore A{B}^2=A{C}^2-C{D}^2+B{D}^2\therefore A{B}^2=A{C}^2-{\left(\frac{1}{4}BC\right)}^2+{\left(\frac{3}{4}BC\right)}^2\therefore A{B}^2=A{C}^2-\frac{B{C}^2}{16}+\frac{9B{C}^2}{16}\therefore A{B}^2=A{C}^2-\frac{B{C}^2}{16}+\frac{9B{C}^2}{16}\therefore A{B}^2=A{C}^2-\left(\frac{B{C}^2-9B{C}^2}{16}\right)\therefore A{B}^2=A{C}^2+\frac{\text{/}{8}^{1}B{C}^2}{\text{/}{16}_2}\therefore A{B}^2=A{C}^2+\frac{B{C}^2}{2}\therefore 2A{B}^2=2A{C}^2+B{C}^2$
Hence proved