Question

For Exercises 23-27 write proofs in paragraph form. (Hint: you can use theorems from this section to write fairly short proofs for exercise 23 and 24.)

Given: $\mathbf{m}\angle \mathbf{RTS}\mathbf{=}\mathbf{90}$;$\overleftrightarrow{\mathrm{MN}}$ is the$\perp$ bisector of $\overline{\mathrm{TS}}$.

Prove:$\overline{\mathrm{TM}}$is a median.

Short answer

It is being given that $\mathbf{m}\angle \mathbf{RTS}\mathbf{=}\mathbf{90}$; $\overleftrightarrow{\mathrm{MN}}$is the$\perp$ bisector of $\overline{\mathrm{TS}}$.

As, $\overleftrightarrow{\mathrm{MN}}$is the$\perp$ bisector of $\overline{\mathrm{TS}}$, therefore$\angle \mathbf{MNS}\mathbf{=}\mathbf{90}$ and $\mathbf{TN}\mathbf{=}\mathbf{NS}$.

In the triangles$\mathrm{\Delta }\mathbf{RTS}$ and $\mathrm{\Delta }\mathbf{MNS}$, it can be noticed that:

$\angle \mathbf{RTS}\mathbf{=}\angle \mathbf{MNS}\mathbf{=}\mathbf{90}$

$\angle \mathbf{RST}\mathbf{=}\angle \mathbf{MSN}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\left(\mathrm{common}\right)$

Therefore, the triangles$\mathrm{\Delta }\mathbf{RTS}$ and$\mathrm{\Delta }\mathbf{MNS}$are similar triangles.

In the similar triangles, the corresponding sides are in proportion.

Therefore, it can be obtained that:

$$\begin{gathered} \frac{\mathrm{RS}}{\mathrm{MS}}\mathbf{=}\frac{\mathrm{TS}}{\mathrm{NS}} \\ \frac{\mathrm{RS}}{\mathrm{MS}}\mathbf{=}\frac{\mathrm{TN}+\mathrm{NS}}{\mathrm{NS}} \\ \frac{\mathrm{RS}}{\mathrm{MS}}\mathbf{=}\frac{\mathrm{NS}+\mathrm{NS}}{\mathrm{NS}} \\ \frac{\mathrm{RS}}{\mathrm{MS}}\mathbf{=}\frac{2\mathrm{NS}}{\mathrm{NS}} \\ \frac{\mathrm{RS}}{\mathrm{MS}}\mathbf{=}\mathbf{2} \\ \mathbf{RS}\mathbf{=}\mathbf{2}\mathbf{MS} \\ \mathbf{RM}\mathbf{+}\mathbf{MS}\mathbf{=}\mathbf{2}\mathbf{MS} \\ \mathbf{RM}\mathbf{=}\mathbf{2}\mathbf{MS}\mathbf{-}\mathbf{MS} \\ \mathbf{RM}\mathbf{=}\mathbf{MS} \end{gathered}$$

As $\mathbf{RM}\mathbf{=}\mathbf{MS}$, therefore it can be obtained that $\mathbf{M}$is the midpoint of $\mathbf{TS}$.

Therefore,$\mathbf{TM}$ is the median.

Hence proved.

Step-by-step solution

Step 1. Observe the given diagram.

The given diagram is:

Step 2. Description of step.

It is being given that$m\angle RTS=90$; $\overleftrightarrow{MN}$is the$\perp$ bisector of $\overline{TS}$.

As, $\overleftrightarrow{MN}$is the$\perp$ bisector of $\overline{TS}$, therefore$\angle MNS=90$ and $TN=NS$.

In the triangles$\mathrm{\Delta }RTS$ and $\mathrm{\Delta }MNS$, it can be noticed that:

$\angle RTS=\angle MNS=90$

$\angle RST=\angle MSN\left(\mathrm{common}\right)$

Therefore, the triangles$\mathrm{\Delta }RTS$ and$\mathrm{\Delta }MNS$are similar triangles.

Step 3. Description of step.

In the similar triangles, the corresponding sides are in proportion.

Therefore, it can be obtained that:

$$\begin{gathered} \frac{RS}{MS}=\frac{TS}{NS} \\ \frac{RS}{MS}=\frac{TN+NS}{NS} \\ \frac{RS}{MS}=\frac{NS+NS}{NS} \\ \frac{RS}{MS}=\frac{2NS}{NS} \\ \frac{RS}{MS}=2 \\ RS=2MS \\ RM+MS=2MS \\ RM=2MS-MS \\ RM=MS \end{gathered}$$

As $RM=MS$, therefore it can be obtained that $M$is the midpoint of $RS$.

Therefore,$TM$is the median.

Hence proved.