Question

In the adjoining figure, if $BC=aunits,AC=bunits,AB=cunits$ and $\angle CAB={120}^{\circ }$, then prove that $a^2$ = $b^2$ + $c^2$ + bc.

Answer 1

In $△CDB$,

${BC}^2={CD}^2+{BD}^2\text{[Pythagoras theorem]}$

${BC}^2={CD}^2+{(DA+AB)}^2$

${BC}^2={CD}^2+{DA}^2+{AB}^2+(2\times DA\times AB)...\left(i\right)$

In $△ADC$,

${CD}^2+{DA}^2={AC}^2...\left(ii\right)\text{[Pythagoras Theorem]}$

Here, $\angle CAB={120}^{\circ }$ (given)

$\Rightarrow \angle CAD={60}^{\circ }$ (since $\angle CAD$ and $\angle CAB$ form a linear pair of angles)

Also, $\cos {60}^{\circ }=\frac{AD}{AC}$

$AC=2AD...\left(iii\right)$

Substituting the values from $\left(ii\right)\&\left(iii\right)\text{in}\left(i\right)$ we get,

${BC}^2={AC}^2+{AB}^2+(AC\times AB)$

$a^2=b^2+c^2+bc$