complete the flow proof.
given: $ \overline { c t } \cong \overline { z m } $,
$ \overline { c m } \perp \overline { t k } $, $ \overline { z k } \perp \overline { t k } $,
$ m $ is the midpoint of $ \overline { t k } $
prove: $ \triangle c t m \cong \triangle z m k $
$ \overline { c m } \perp \overline { t k } $ $ \overline { z k } \perp \overline { t k } $
definition of perpendicular lines
definition of a right triangle
given
hypotenuse-leg theorem
definition of a midpoint

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Accepted Answer

# Explanation:
## Step1: Use the definition of perpendicular lines
Since $\overline{CM}\perp\overline{TK}$ and $\overline{ZK}\perp\overline{TK}$, by the definition of perpendicular lines, $\angle CMT = 90^{\circ}$ and $\angle ZKM=90^{\circ}$

## Step2: Use the mid - point definition
Given $M$ is the midpoint of $\overline{TK}$, by the definition of a midpoint, $TM = MK$

## Step3: Use the Hypotenuse - Leg Theorem
We are given $\overline{CT}\cong\overline{ZM}$, and from Step1 $\angle CMT=\angle ZKM = 90^{\circ}$, from Step2 $TM = MK$. In right - triangles $\triangle CTM$ and $\triangle ZMK$, by the Hypotenuse - Leg ($HL$) Theorem (if the hypotenuse and a leg of one right - triangle are congruent to the hypotenuse and a leg of another right - triangle, then the two right - triangles are congruent), $\triangle CTM\cong\triangle ZMK$

# Answer:
$\triangle CTM\cong\triangle ZMK$ by the Hypotenuse - Leg ($HL$) Theorem.

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QUESTION IMAGE

Question

Explanation:

Step1: Use the definition of perpendicular lines

Step2: Use the mid - point definition

Step3: Use the Hypotenuse - Leg Theorem

Answer: