Question

• $overleftrightarrow{bd}$ is the perpendicular bisector of segment $overline{ac}$. • $overline{bd}$ is 3 units long. • $overline{ac}$ is 8 units long. naomi was asked to show that point $d$ is equidistant from points $a$ and $c$. image of a coordinate-like diagram with points a, b, c, d where b is the midpoint of ac, d is on the perpendicular line through b, and a right angle at b between ac and bd fill in the blanks in naomi’s solution. • since $overleftrightarrow{bd}$ bisects $overline{ac}$, we know that both $overline{ab}$ and $overline{bc}$ are $square$ units long. • we are given that $overline{bd}$ is 3 units long. • $overleftrightarrow{bd}$ is perpendicular to $overline{ac}$, so $\triangle abd$ is a right triangle. • applying the pythagorean theorem to $\triangle abd$ shows that the length of the hypotenuse is $square$ units. • we can follow a similar process on $\triangle cbd$ to show that $square$ have the same length, so point $d$ must be equidistant from points $a$ and $c$.

Explanation:

Step1: Find length of AB and BC

Since \( \overleftrightarrow{BD} \) bisects \( \overline{AC} \), \( AB = BC=\frac{AC}{2} \). Given \( AC = 8 \) units, so \( AB=BC=\frac{8}{2}=4 \) units.

Step2: Apply Pythagorean theorem to \( \triangle ABD \)

In right triangle \( \triangle ABD \), \( AB = 4 \) units, \( BD = 3 \) units. By Pythagorean theorem \( AD=\sqrt{AB^{2}+BD^{2}}=\sqrt{4^{2}+3^{2}}=\sqrt{16 + 9}=\sqrt{25}=5 \) units.

Step3: Analyze \( \triangle CBD \)

For \( \triangle CBD \), \( BC = 4 \) units, \( BD = 3 \) units. Using Pythagorean theorem, \( CD=\sqrt{BC^{2}+BD^{2}}=\sqrt{4^{2}+3^{2}} = 5 \) units. So \( AD = CD \), meaning \( AD \) and \( CD \) have the same length.

Answer:

• First blank: \( 4 \)
• Second blank: \( 5 \)
• Third blank: \( AD \) and \( CD \)