Mathematics - Properties of Triangle Question with Solution | TestHub

Applying cosines rule in $△ABD$ and $△BCD$, we get $2a^2+2b^2=d_1^2+d_2^2$, where ${\mathrm{d}}_1,{\mathrm{d}}_2$ are length of diagonals $\left({\mathrm{d}}_1<{\mathrm{d}}_2\right)$ given ${\mathrm{d}}_2^2=3{\mathrm{d}}_1^2$

$\Rightarrow {\mathrm{a}}^2+{\mathrm{b}}^2=2{\mathrm{d}}_1^2$ also $ab=d_1^2..\dots \dots \dots \dots \left(1\right)$

$$\Rightarrow a^2+b^2+2ab=3ab+d_1^2=4d_1^2..\dots \dots \dots \dots \left(2\right)$$

$$\Rightarrow (\mathrm{a}+\mathrm{b}{)}^2=4{\mathrm{d}}_1^2\Rightarrow \mathrm{a}+\mathrm{b}=2{\mathrm{d}}_1$$

using (1) and (2), we get $\mathrm{a}=\mathrm{b}={\mathrm{d}}_1$ i.e. $\frac{\mathrm{a}}{\mathrm{b}}=1$.