Question

Prove that the area of the triangle described on one side $BC$ of a square $ABCD$ as base is half the area of the similar triangle $ACF$ produced on the diagonal $AC$ as base.

Answer 1

ABCD is a square with AC as its diagonal.

Then,

$A{C}^2=A{B}^2+B{C}^2=B{C}^2+B{C}^2=2B{C}^2$

$\Rightarrow AC=\sqrt{2}BC$

Also, it is given that $\Delta$BCE $\sim \Delta$ ACF

Using the theorem, "The area of $2$ similar triangles are in ratio of squares of the corresponding sides."

$\frac{Area\left(\Delta BCF\right)}{Area\left(\Delta ACF\right)}=\frac{B{C}^2}{A{C}^2}=\frac{B{C}^2}{(\sqrt{2}BC{)}^2}=\frac{1}{2}$

$\Rightarrow$ Area $\left(\Delta BCE\right)=\frac{1}{2}$.

Area $\left(\Delta ACF\right)$

Hence, the area of $\Delta$BCE is half of the area of $\Delta$ACF.