Show that the diagonal of a rectangle divides it in two congruent triangles.

In the given figure, if $DE\parallel BC, AD=2.5 cm, DB=3.5 cm$ and $EC=4.2 cm$, then the measure of $AC$ is:

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 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 $ABC$ ?