91Ó°ÊÓ

$\mathrm{Δ}XYZ$is:

Here, M, N and T are midpoints of $\mathrm{Δ}XYZ$. And $XY=k$.

Consider in triangles TXM and ZXY .

$∠TXM=∠ZXY$ (Common).

$\frac{TX}{ZX}=\frac{XM}{XY}=\frac{1}{2}$ (Midpoints).

Therefore, $\mathrm{Δ}TXM~\mathrm{Δ}ZXY$(SAS).

$\begin{array}{l}∠XMT=∠XYZ\\ ∠MTX=∠YZX\end{array}$

Angle XMT and XYZ are the corresponding angles. Hence, line TM and $ZY$ or ZN are parallel to each other.

Consider in triangles NMY and ZXY .

$∠NYM=∠ZYX$ (Common).

$\frac{NY}{ZY}=\frac{YM}{YX}=\frac{1}{2}$ (Midpoints).

Therefore, $\mathrm{Δ}NMY~\mathrm{Δ}ZXY$(SAS).

$\begin{array}{l}∠YNM=∠YZX\\ ∠YMN=∠YXZ\end{array}$

Angle YNM and YZX are the corresponding angles. Hence, line NM and ZX or ZT are parallel to each other.

Thus, MNZT is parallelogram.

Similarly, TNXM is also a parallelogram

$TN=XM$.

The side TN is calculated as follows:

$\begin{array}{c}TN=XM\\ =\frac{XY}{2}\\ TN=\frac{k}{2}\text{units}\end{array}$

Therefore, the length of TN is $\frac{k}{2}$units.