From any point on the bisector of an angle of a triangle, perpendiculars are drawn to the arms of the angle, prove that the perpendiculars are equal. 

In a triangle $ABC$ with $\angle A<\angle B<\angle C,$ points $D,E,F$ are on the interior of segments $BC,CA,AB$ respectively. Which of the following triangles cannot be similar to the triangle $ABC$ ?

In a quadrilateral $ACBD, AC=AD$ and $AB$ bisects $\angle A$(see figure). Show that $△ABC\cong △ABD$. What can you say about $BC$ and $BD?$     

Let $ABC$ be a triangle and $M$ be a point on side $AC$ closer to vertex $C$ than $A$. Let $N$ be a point on side $AB$ such that $MN$ is parallel to $BC$ and let $P$ be a point  on side $BC$ such that $MP$ is parallel to $AB$. If the area of the quadrilateral $BNMP$ is equal to $\frac{5}{18}$ of the area of $\mathrm{\Delta }ABC$, then the ratio $AM/MC$ equals

In the figure, in $△ABC, AB=AC=10 \text{cm} \text{and} BC=12 \text{cm}.$ $P \text{and} Q$ are the midpoints of $AB \text{and} AC$ respectively. $PM \text{and} RN$ are perpendiculars on $SQ$. If $BS : SR : RC=1 : 2 : 1$, then the length of $MN$ is:

In the figure given below, $M$ is the mid-point of $AB \text{and} \angle DAB=\angle CBA\text{ and} \angle AMC=\angle BMD$. Then the triangle $ADM$ is congruent to the triangle $BCM$ by______.

In the figure $ABC$ is an equilateral triangle with side $14$ cm. $AX=\frac{1}{3}AB, BY=\frac{1}{3}BC \text{and} CZ=\frac{1}{3}AC$. What is the area (in ${\mathrm{cm}}^2$ ) of $△PQR$?