91Ó°ÊÓ

Consider the diagram.

Consider the triangle$â–³RST$.

Here, points Mand N are the mid points of$\stackrel{¯}{RT}$and $\stackrel{¯}{RS}$.

So, the side $\stackrel{¯}{MN}$is half as long as the third side$\stackrel{¯}{ST}$.

$\begin{array}{c}MN=\frac{1}{2}ST\\ =\frac{1}{2}(SE+ET)\\ =\frac{1}{2}(18+10)\end{array}$

$\begin{array}{c}=\frac{1}{2}\left(28\right)\\ =14\end{array}$

Consider the triangle $â–³RST$.

It is given that $\stackrel{¯}{RE}$ is an altitude of $△RST$.

So, it can be said that the triangles $â–³SER$and $â–³TER$ are right triangles.

Now,$\stackrel{¯}{RS}$ is the hypotenuse of the right triangle $△SER$ and N is its midpoint.

So, the midpoint N of the hypotenuse $\stackrel{¯}{RS}$ of a right triangle $△SER$ is equidistant from the three vertices.

Thus,$NE=NR=15$ .

Similarly,$\stackrel{¯}{RT}$ is the hypotenuse of the right triangle$△TER$ and M is its midpoint.

So, the midpoint N of the hypotenuse $\stackrel{¯}{RT}$ of a right triangle $△TER$is equidistant from the three vertices.

Thus, $MT=ME=13$.

Also, M is the midpoint of $\stackrel{¯}{RT}$. So, it implies that,$RM=MT=13$ .

Therefore,

$\begin{array}{c}RT=RM+MT\\ =13+13\\ =26\end{array}$

Therefore,$MN=14,RT=26,NE=15$ .