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bisect a segment to find the midpoint. two segments that have the same …

Question

bisect a segment to find the midpoint. two segments that have the same length are congruent and you can write a congruence statement using the symbol ≅. so, if the segments $overline{pq}$ and $overline{rs}$ are the same length, you can write $overline{pq}congoverline{rs}$. the tick marks on $overline{pq}$ and $overline{rs}$ indicate that the segments are congruent. multiple tick marks may be used, but two segments must have the same number of tick marks to indicate that the segments are congruent. to bisect a figure is to divide it into two congruent parts. the mid - point of a segment is the point that bisects the segment into two congruent segments. this task comes from on your own numbers 13 and 15 and is used as a substitute for the construction. task 3: a. the length of $overline{df}$ is $(4x + 2)$ inches, with $e$ as the midpoint. the length of $overline{de}$ is 17 inches. what is the value of $x$? b. point $b$ lies along $overline{ac}$ between points $a$ and $c$. the length of $overline{ac}$ is 38 centimeters. if $ab = 7x-1$ and $bc = 4x + 6$, is $b$ the midpoint of $overline{ac}$? explain your reasoning

Explanation:

Step1: Recall mid - point property

Since \(E\) is the mid - point of \(\overline{DF}\), then \(DE=\frac{1}{2}DF\).

Step2: Substitute given values

We know \(DF = 4x + 2\) and \(DE=17\). So, \(17=\frac{4x + 2}{2}\).

Step3: Solve the equation for \(x\)

Multiply both sides of the equation by 2: \(17\times2=4x + 2\), which gives \(34 = 4x+2\). Subtract 2 from both sides: \(34 - 2=4x\), so \(32 = 4x\). Divide both sides by 4: \(x=\frac{32}{4}=8\).

Step4: For part B, use mid - point condition

If \(B\) is the mid - point of \(\overline{AC}\), then \(AB = BC\) and \(AB+BC=AC\). First, set \(AB = BC\): \(7x−1=4x + 6\). Subtract \(4x\) from both sides: \(7x-4x-1=4x-4x + 6\), which gives \(3x-1 = 6\). Add 1 to both sides: \(3x=7\), so \(x=\frac{7}{3}\). Then find \(AB = 7\times\frac{7}{3}-1=\frac{49}{3}-1=\frac{49 - 3}{3}=\frac{46}{3}\) and \(BC=4\times\frac{7}{3}+6=\frac{28}{3}+6=\frac{28 + 18}{3}=\frac{46}{3}\). Also, \(AB + BC=\frac{46}{3}+\frac{46}{3}=\frac{92}{3}
eq38\). So \(B\) is not the mid - point.

Answer:

A. \(x = 8\)
B. \(B\) is not the mid - point of \(\overline{AC}\) because when we solve for \(x\) using \(AB = BC\), we get \(x=\frac{7}{3}\), and \(AB + BC
eq AC\).