Question

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

Explanation:

Step1: Definition of perpendicular lines

If two lines are perpendicular, then they form right angles. Since \(\overline{CM}\perp\overline{TK}\) and \(\overline{ZK}\perp\overline{TK}\), we have \(\angle CMT = 90^{\circ}\) and \(\angle ZKM=90^{\circ}\)

Step2: Definition of a midpoint

Given that \(M\) is the mid - point of \(\overline{TK}\), by the definition of a mid - point, \(TM = MK\)

Step3: Hypotenuse - Leg Theorem

We are given that \(\overline{CT}\cong\overline{ZM}\). In right triangles \(\triangle CTM\) and \(\triangle ZMK\), we have the hypotenuse \(CT\cong ZM\) (given) and the leg \(TM\cong MK\) (from the mid - point definition). By the Hypotenuse - Leg (\(HL\)) Theorem for right - triangle congruence, \(\triangle CTM\cong\triangle ZMK\)

Answer:

The blanks in the flow - proof are filled as follows: First blank (for \(\overline{CM}\perp\overline{TK},\overline{ZK}\perp\overline{TK}\)): Definition of perpendicular lines; Second blank (for \(M\) is the mid - point of \(\overline{TK}\)): Definition of a midpoint; Third blank (for proving \(\triangle CTM\cong\triangle ZMK\)): Hypotenuse - Leg Theorem.