Prove that the area of the equilateral triangle described on the side of a square is half the area of the equilateral triangles described on its diagonal.

$ABC$ is a triangle and $PQ$ is a straight line meeting $AB$ in $P$ and $AC$ in $Q$. If $AP=1 \mathrm{cm}, PB=3 \mathrm{cm},\mathit{ }AQ=1.5 \mathrm{cm}, QC=4.5 \mathrm{m}$, prove that area of $△APQ$ is one sixteenth of the area of $△ABC$.  

In the given figure, $PS, SQ, PT$ and $TR$ are $4 \mathrm{cm}, 1 \mathrm{cm}, 6 \mathrm{cm}$ and $1.5 \mathrm{cm}$, respectively. Prove that $ST\parallel QR$. Also, find $\frac{ar \left(△PST\right)}{ar (\mathrm{trapezium} QRTS)}$.

In the given figure, in $△ABC, XY\parallel AC$ and $XY$ divides the $△ABC$ into two regions such that $ar \left(△BXY\right)=2ar \left(ACYX\right)$. Determine $\frac{AX}{AB}$.

In the given figure, $ABC$ is a right-angled triangle with $\angle BAC=90°$. Find the ratio of the area of $△ADB$ to the area of $△CDA$.

In the given figure, if $\mathrm{DE}\parallel \mathrm{BC}$ and $\mathrm{AD} : \mathrm{DB}=5 : 4$, then find $\frac{\mathrm{ar} \left(△\mathrm{DFE}\right)}{\mathrm{ar} \left(△\mathrm{CFB}\right)}$.

In the given figure, $\angle PQR=\angle PST=90°, PQ=5 \mathrm{cm}$ and $PS = 2 \mathrm{cm}$. Find Area of $△PQR :$ Area of quadrilateral $SRQT$.

In the figure, the perpendicular from $A$ on side $BC$ of a $△ABC$, intersects $BC$ at $D$ such that $DB=3CD$. Prove that $2A{B}^2=2A{C}^2+B{C}^2$.