Statement-I    In $\Delta ABC,AD\perp BC$  and $A{D}^2=BD\times CD$ , then $\angle A=90°$ .

$\mathbf{Statement}\mathbf{-}\mathbf{II}\mathbf{ } \mathrm{In} \mathrm{an} \mathrm{equilateral} \mathrm{triangle} \mathrm{ABC} \mathrm{the} \mathrm{side} \mathrm{BC} \mathrm{is} \mathrm{trisected} \mathrm{at} \mathrm{D}, \mathrm{then} 9{\mathrm{AD}}^2=7{\mathrm{AB}}^2.$

Statement A:

$A{D}^2=BD\times CD$  then

$\angle A=90°$   (True)

Statement B:

$BD=\frac{a}{3}$

$BE=\frac{a}{2}$   $\left[\because BE\text{is median and }BE\perp BC\right]$

$DE=BE-BD$

$=\frac{a}{2}-\frac{a}{3}=\frac{3a-2a}{6}=\frac{a}{6}$

$\therefore DE=\frac{a}{6}$

From $\Delta ABE$ ,

$A{B}^2=B{E}^2+A{E}^2\to \left(1\right)$

$A{B}^2={\left(\frac{a}{2}\right)}^2+A{E}^2=\frac{a^2}{4}+A{E}^2$

From $\Delta ADE$ ,

$A{D}^2=A{E}^2+E{D}^2\to \left(2\right)$

From (1) and (2)

$A{B}^2=A{D}^2-E{D}^2+E{B}^2$

$\Rightarrow A{B}^2=A{B}^2-\frac{B{C}^2}{36}+\frac{B{C}^2}{4}$

$\left[\because BD+DE=\frac{BC}{2}\Rightarrow \frac{BC}{3}+DE=\frac{BC}{2}\Rightarrow DE=\frac{BC}{6}\right]$

$\Rightarrow A{D}^2=A{B}^2+\frac{B{C}^2}{36}-\frac{B{C}^2}{4}$

$=36A{B}^2+A{B}^2-9A{B}^2$

$\left[\because BC=AB\right]$

$=\frac{7A{B}^2}{9}$

$\Rightarrow A{D}^2=\frac{7A{B}^2}{9}$

$\Rightarrow 9A{D}^2=7A{B}^2$