Chapter 4.5, Problem 5ST2

To determine

To prove: $\overline{CM}\cong \overline{BN}$

Explanation of Solution

Given information: $\overline{MB}\cong \overline{NC},\overline{BN}\perp \overline{AC},\overline{CM}\perp \overline{AB}$

Proof:

  

First prove $\Delta BMC\cong \Delta CNB$

  $\overline{MB}\cong \overline{NC}$ (Given)

As $\overline{BN}\perp \overline{AC}$ and $\overline{CM}\perp \overline{AB}$ ,

  $m\angle BMC=90°,m\angle CNB=90°$

So,

  $\angle BMC\cong \angle CNB$ (Right angles are congruent to each other)

  $\overline{BC}\cong \overline{BC}$ (Reflexive property of congruence)

(According to reflexive property of congruence, a line segment is congruent to itself)

Thus,

  $\Delta BMC\cong \Delta CNB$ (RHS congruence criteria)

(According to RHS congruence criteria, two right angled triangles are said to be congruent if hypotenuse and one side of one triangle are congruent to the hypotenuse and one side of another triangle)

Therefore,

  $\overline{CM}\cong \overline{BN}$ (Corresponding parts of congruent triangles)