Chapter 4.2, Problem 21WE

To determine

To prove: $\Delta POA\cong \Delta POB$ using two column form.

Explanation of Solution

Given information:

Given figure,

Given:

Plane $M$ bisects $\overline{AB};$

  $\overline{PA}\perp \overline{AB}$

Calculation:

SSS Postulate: if there sides of triangle are equal to the three corresponding sides of another triangle, then the triangles are congruent.

SAS Postulate: If two sides and the angle included between them of triangle are equal to the two corresponding sides and the angle included between them of another triangle, then the triangles are congruent.

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

Given:

Plane $M$ bisects $\overline{AB};$

  $\overline{PA}\perp \overline{AB}$

Prove: $\Delta POA\cong \Delta POB$

Proof:

Statements	Reasons
1. $M$ bisects   $\overline{AB}$	Given
2. $\overline{AO}\cong \overline{BO}$	Definition of midpoint
3. $\overline{PA}\perp \overline{AB}$	Given
4. $m\angle POA=90°;m\angle POB=90°$	Definition of perpendicular lines
5. $\Delta POA\cong \Delta POB$	Definition of congruent angles
6. $\overline{PO}\cong \overline{PO}$	Reflexive property
7. $\Delta POA\cong \Delta POB$	SAS Postulate