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 $c$. 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 BD bisects AC, AB = BC = AC/2. AC is 8 units, so AB = 8/2 = 4.

Step2: Apply Pythagorean theorem to ΔABD

In right triangle ABD, legs are AB = 4 and BD = 3. Hypotenuse AD: \( AD = \sqrt{AB^2 + BD^2} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \).

Step3: Analyze ΔCBD

Similarly, in ΔCBD (right triangle), CB = 4, BD = 3. So CD = \( \sqrt{4^2 + 3^2} = 5 \), so AD and CD have same length.

Answer:

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