Question

please answer that 18th question please ​

Answer 1

Know that

• $D/4=0$
• $D/4=(ac+bd)^2-(a^2+b^2)(c^2+d^2)$

$(ac+bd)^2=a^2c^2+2abcd+b^2d^2$

$-(a^2+b^2)(c^2+d^2)=-(a^2c^2+a^2d^2+b^2c^2+b^2d^2)$

$-a^2d^2+2abcd-b^2c^2=0$

$a^2d^2-2abcd+b^2c^2=0$

$(ad-bc)^2=0$

∴ $ad=bc$

Hence, $a:b=c:d$.

We know that $a:b=\frac{a}{b}$ and $c:d=\frac{c}{d}$.

∴ $\frac{a}{b} =\frac{c}{d}$

Proved.