Question

In the given figure, AD ⊥ BC and $\mathrm{BD}=\frac{1}{3}\mathrm{CD}.$
Prove that $2C{A}^2=2A{B}^2+B{C}^2.$

Answer 1

Given: In ΔABC, AD⊥BC and $\mathrm{BD}=\frac{1}{3}\mathrm{CD}$.
To prove: 2CA² = 2AB² + BC²
Proof:
$\mathrm{BD}=\frac{1}{3}\mathrm{CD}$ ............(i)
∴ BC = BD + CD
$\mathrm{BC}=\frac{1}{3}\mathrm{CD}+\mathrm{CD}=\frac{4}{3}\mathrm{CD}$
or,$CD=\frac{3}{4}BC$
and $BD=\frac{1}{3}CD=\frac{1}{3}\times \frac{3}{4}BC=\frac{1}{4}BC$
$$\begin{gathered} \mathrm{In}\Delta \; ADC,\angle \mathrm{AD}C=\; 90°. \\ A{C}^2=A{D}^2+D{C}^2...\left(\mathrm{i}\right)(\mathrm{By}\text{P}\mathrm{ythagoras}\text{'}\text{t}\mathrm{heroem}) \\ \mathrm{In}\Delta \; ADB,\angle \mathrm{ADB\; =\; 90}°. \\ A{B}^2=A{D}^2+D{B}^2 \\ \Rightarrow A{D}^2=A{B}^2-D{B}^2 \\ BysubstitutingthevalueofA{D}^{2}in\left(i\right)weget: \\ A{C}^2=A{B}^2-D{B}^2+D{C}^2 \\ AgainsubstitutingthevaluesofDBandDC,weget: \\ \Rightarrow A{C}^2=A{B}^2-\frac{1}{16}B{C}^2+\frac{9}{16}B{C}^2 \\ \Rightarrow A{C}^2=A{B}^2+\left(\frac{9}{16}-\frac{1}{16}\right)B{C}^2 \\ \Rightarrow A{C}^2=A{B}^2+\frac{1}{2}B{C}^2 \\ \Rightarrow 2A{C}^2=2A{B}^2+B{C}^2 \end{gathered}$$