Similarity of Triangles

In $∆DEW, AB \left|\right| EW$. If $AD = 4 \mathrm{cm}$, $DE = 12 \mathrm{cm}$ and $DW = 24 \mathrm{cm}$, then find the value of $DB$.

In the given figure, find $\mathrm{PT}$ given that $l_1\parallel l_2$

In $△ ABC, DE\parallel DE\parallel BC$. If $AD=5 \mathrm{cm}, BD=7 \mathrm{cm}$ and $AC=18 \mathrm{cm}$, find the length of $AE$ in centimrter.

$ABCD$ is a trapezium in which $AB\parallel DC$ and its diagonals intersect each other at the point $O$, then which of the following is correct:

In the figure, if $△POQ~△SOR$ and $PQ : RS=$ $1:2,$ then $OP:OS$ is 

In $△PQR, NM\parallel RQ$. If $PM=15, MQ=10, NR=8$, then find $PN$.

In $∆ABC$, $DE\parallel BC, AD=6 \mathrm{cm}, DB=9 \mathrm{cm}$ and $AE=8 \mathrm{cm}$, then measure of $AC$ (in $\mathrm{cm}$) will be:

The points $D$ and $E$ are points on the sides $AB$ and $AC$ in a $△ABC$, such that $DE\parallel BC$. If $AD=4x-3, AE=8x-7, BD=3x-1$ and $CE=5x-3$, then find the value of $x$.

Two isosceles triangles are similar, if an angle of one is congruent to the corresponding angle of the other.

All isosceles triangles are similar.

In the figure, find the locus of the points inside the $\angle \mathrm{ABC}$ and equidistant from two sides $\mathrm{BA}$ and $\mathrm{BC}$.

In the given figure, $\mathrm{PS}$ is the bisector of the angle $\angle \mathrm{P}$ intersect $\mathrm{QR}$ at $\mathrm{S}$. $\mathrm{SN}\perp \mathrm{PQ}$ and $\mathrm{SM}\perp \mathrm{PR}$ are drawn. Prove that $\mathrm{SN}=\mathrm{SM}$.

If $△ABC~△DEA$, also $D$ and $E$ points on the lines $AB \mathrm{and} AC$ such that $DE\parallel BC$ and $AD=8 \mathrm{cm}, AB=12 \mathrm{cm}$ and $AE=12 \mathrm{cm}$, the measure of $CE$ is

In the figure, if $DE\parallel BC$ the value of $x$ is:

If a perpendicular is drawn from the vertex $A$ to the opposite side$BC$, we get $A{D}^2=BD\times DC$, then prove that $∆ABC$ is a right-angled triangle.

In $△ABC$ and $△FDE$, state whether $\frac{AB}{DE}=\frac{BC}{EF}=\frac{CA}{FD}$ is true or not? Justify your answer.

In trapezium $ABCD$, diagonals intersects each other at $O$ and $AB\parallel DC$. Show that$\frac{ AO}{BO}=\frac{CO}{DO}$.

In the given figure, if $EF\parallel DC\parallel AB$, then prove that $\frac{AE}{ED}=\frac{BF}{FC}$.

In the figure, if $DE\parallel BC$ and $CD\parallel EF$, then prove $A{D}^2=AB\times AF$.

In $△ABC$, points $D$ and $E$ are respectively situated on the sides $AB$ and $BC$ such that $BD=CE$. If $\angle B=\angle C$, then show that $DE\parallel BC$.