Chapter 5.7, Problem 18PSB

a

To determine

To prove: $PMQC$ is a parallelogram.

Explanation of Solution

Given information:

In $\Delta ABC$ ,

  $M$ is the mid-point of $\overline{AB}$ .

  $\overline{MQ}\text{ is parallel to }\overline{\text{AC}}$

  $\overline{MP}\text{ is parallel to }\overline{\text{QC}}$

Formula used:

If opposite sides of quadrilateral are parallel, then it is a parallelogram.

Proof:

In $PMQC$ ,

  $\overline{MQ}\parallel \overline{\text{AC}}$    [ $\because$ Given]

  $\overline{MP}\parallel \overline{\text{QC}}$    [ $\because$ Given]

Since, opposite sides of quadrilateral $\left(PMQC\right)$ are parallel, then $PMQC$ is a parallelogram

b

To determine

To prove: $\Delta MAP\text{ is congruent to }\Delta \text{BMQ}$

Explanation of Solution

Given information:

In $\Delta ABC$ ,

  $M$ is the mid-point of $\overline{AB}$ .

  $\overline{MQ}\text{ is parallel to }\overline{\text{AC}}$

  $\overline{MP}\text{ is parallel to }\overline{\text{QC}}$

Formula used:

ASA Rule:If two angles and the included side of one triangle are equal to the corresponding angles and side of another triangle, the triangles are congruent.

Proof:

From, $\Delta MAP\text{ and }\Delta \text{BMQ}$

  $\angle AMP=\angle MBQ$    [ $\because$ Corresponding Angles]

  $AM=MB$   [ $\because$ Given $M$ is the mid-point of $\overline{A\text{}B}$ ]

  $\angle MAP=\angle BMQ$    [ $\because$ Corresponding Angles]

Therefore, from ASA rule $\Delta MAP\cong \Delta \text{BMQ}$