Chapter 5.3, Problem 5CE

To determine

To find: The length of TN .

Answer to Problem 5CE

The length of TN is $\frac{k}{2}$ units.

Explanation of Solution

Given:

The triangle is shown below:

  

Side XY is k units. M , N and T are the midpoints.

Calculation:

Consider in triangles TXM and ZXY .

  $\begin{array}{l}\angle TXM=\angle ZXY\left(\text{Common}\right)\\ \frac{TX}{ZX}=\frac{XM}{XY}=\frac{1}{2}\left(\text{Midpoints}\right)\\ \therefore \Delta ABF\sim \Delta CDE\left(\text{SAS}\right)\\ \angle XMT=\angle XYZ\\ \angle MTX=\angle YZX\end{array}$

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 .

  $\begin{array}{l}\angle NYM=\angle ZYX\left(\text{Common}\right)\\ \frac{NY}{ZY}=\frac{YM}{YX}=\frac{1}{2}\left(\text{Midpoints}\right)\\ \therefore \Delta NMY\sim \Delta ZXY\left(\text{SAS}\right)\\ \angle YNM=\angle YZX\\ \angle YMN=\angle YXZ\end{array}$

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:

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

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