Quadrilaterals — Maths STD 9 — Question

4. BC² + AD² + 2AB ⋅ CD

Solution:

Construction: Draw perpendicular from D and C on AB which meets AB at E and F, respectively.

So, DEFC is a parallelogram, since one pair of opposite sides are parallel and equal.

In $\triangle\text{ABC},$

$\angle\text{B}$ is an acute angle.

$\Rightarrow\text{AC}^2=\text{BC}^2+\text{AB}^2-2\text{AB}\times\text{AE}$

In $\triangle\text{ABD},$

$\angle\text{A}$ is an acute angle.

$\Rightarrow\text{BD}^2=\text{AD}^2+\text{AB}^2-2\text{AB}\times\text{AF}$

$\Rightarrow\text{AC}^2+\text{BD}^2=\text{BC}^2+\text{AD}^2+2\text{AB}(\text{AB}-\text{BE}-\text{AF})$

$=\text{BC}^2+\text{AD}^2+2\text{AB}\times\text{EF}$

$=\text{BC}^2+\text{AD}^2+2\text{AB}\times\text{CD}$