Question

Write proofs in two-column form.

Given: Plane$M$ bisects $\overline{AB}$;$\overline{PA}\cong \overline{PB}$

Prove:$△POA\cong △POB$

Short answer

The two-column proof is:

Statements	Reasons
$M$bisects$\overline{AB}$	Given
$\overline{AO}\cong \overline{BO}$	Definition of midpoint
$\overline{PA}\cong \overline{PB}$	Given
$\overline{PO}\cong \overline{PO}$	Reflexive property
$△POA\cong △POB$	SSS postulate

Step-by-step solution

Step 1. Observe the figure.

The figure is:

Step 2. Description of step.

Given thatplane$M$bisects $\overline{AB}$ and $PA\cong PB$.

Therefore, by using the definition of midpoint, it can be said that $AO=OB$.

That implies, $AO\cong OB$.

In the triangles $△POA$ and $△POB$, it can be noticed that $PO=PO$ by using the reflexive property.

That implies, $PO\cong PO$.

Therefore, it can be seen that $AO=OB$,$PO=PO$ and $PA=PB$.

Therefore, the triangles $△POA$ and $△POB$ are congruent triangles by using the SSS postulate.

Step 3. Write the proof in two-column form.

The proof in two-column form is:

Statements	Reasons
$M$bisects$\overline{AB}$	Given
$\overline{AO}\cong \overline{BO}$	Definition of midpoint
$\overline{PA}\cong \overline{PB}$	Given
$\overline{PO}\cong \overline{PO}$	Reflexive property
$△POA\cong △POB$	SSS postulate