O P Malhotra, S K Gupta and, Anubhuti Gangal Solutions for Exercise 2: Exercise

The side $BC$ of a square $ABCD$ is produced to any point $E$. Prove that$A{E}^2=2BC.BE+C{E}^2$.

$ABCD$ is a rhombus. Prove that $A{C}^2+B{D}^2=4A{B}^2$

In the figure, $\angle B$ of $△ABC$ is an acute angle and $AD\perp BC$. Prove that $A{C}^2=A{B}^2+B{C}^2-2BC.BD$.

In a quadrilateral $ABCD, \angle B=90° \mathrm{and} A{D}^2=A{B}^2+B{C}^2-C{D}^2$. Prove that$\angle ADC=90°$.

$ABC$ is a triangle right angled at $A$ and $P$ is the length of the perpendicular from $A$ on $BC$. Show that $pa=bc$.

In the figure, $ABC$ is a triangle right angled at $A$ and $p$ is the length of the perpendicular from $A$ on $BC$. Show that $\frac{1}{p^2}=\frac{1}{b^2}+\frac{1}{c^2}$.

In the given figure, $\angle B$ is acute and segment $AD\perp \text{side} BC$. Show that $b^2=h^2+a^2+x^2-2ax$.

In the given figure, $\angle B$ is acute and segment $\mathrm{AD}\perp \mathrm{side} \mathrm{BC}$ Show that $b^2=a^2+c^2-2ax$.