Question

Given: $\Delta DEF$is isosceles with $DF=EF$; $\overline{FX}$bisects$\angle DFE$

a. Would the median drawn from $F$to $\overline{DE}$ be the same segment as $\overline{FX}$?

b. Would the altitude be drawn from $F$to $\overline{DE}$ be the same segment as $\overline{FX}$?

Short answer

a. Yes, the median is drawn from $F$to $\overline{DE}$ be the same segment as $\overline{FX}$

b. Yes, the altitude is drawn from $F$to be $\overline{DE}$ the same segment as$\overline{FX}$

Step-by-step solution

Part a. Step 1. Show that ΔDFX≅ΔEFX

Statement	Reason
1.$\overline{FX}\cong \overline{FX}$	Common line segment
2.$\angle DFX\cong \angle EFX$	Given
3.$\overline{DF}\cong \overline{EF}$	Given
4.$\Delta DFX\cong \Delta EFX$	SAS (Side-Angle-Side)Postulate

Part a. Step 2. Show that DX¯≅EX¯

Since corresponding parts of the congruent triangle are congruent

So,$\overline{DX}\cong \overline{EX}$

Part a. Step 3. Show that FX¯ is a median

Since $\overline{DX}\cong \overline{EX}$, so$X$ is a midpoint of$\overline{DE}$

So, $\overline{FX}$is the median.

Thus, the median drawn from $F$to $\overline{DE}$is the segment$\overline{FX}$

Part b. Step 1. Show that ΔDFX≅ΔEFX

Statement	Reason
1.$\overline{FX}\cong \overline{FX}$	Common line segment
2.$\angle DFX\cong \angle EFX$	Given
3.$\overline{DF}\cong \overline{EF}$	Given
4.$\Delta DFX\cong \Delta EFX$	SAS (Side-Angle-Side)Postulate

Part b. Step 2. Show that ∠FXD≅∠FXE

Since, corresponding parts of the congruent triangle are congruent

So,$\angle FXD\cong \angle FXE$

Part b. Step 3. Show that FX¯ is an altitude

Since $\angle FXD\text{and}\angle FXE$forms a linear pair

Also,$\angle FXD\cong \angle FXE$

$$\begin{gathered} m\angle FXD+m\angle FXE=180° \\ 2m\angle FXD=180° \\ m\angle FXD=90° \end{gathered}$$

So, $\overline{FX}$is the altitude

Thus, the altitude drawn from F to $\overline{DE}$is the segment $\overline{FX}$