Chapter 4: Q19 (page 126)

Write proofs in two-column form.
Given: Plane $M$ bisects $\bar{A B}$ ; $\bar{P A} ≅ \bar{P B}$
Prove: $△ P O A ≅ △ P O B$

Short Answer

Expert verified

The two-column proof is:

Statements	Reasons
$M$ bisects $\bar{A B}$	Given
$\bar{A O} ≅ \bar{B O}$	Definition of midpoint
$\bar{P A} ≅ \bar{P B}$	Given
$\bar{P O} ≅ \bar{P O}$	Reflexive property
$△ P O A ≅ △ P O B$	SSS postulate

Step by step solution

Step 1. Observe the figure.

The figure is:

Step 2. Description of step.

Given thatplane $M$ bisects $\bar{A B}$ and $P A ≅ P B$ .

Therefore, by using the definition of midpoint, it can be said that $A O = O B$ .

That implies, $A O ≅ O B$ .

In the triangles $△ P O A$ and $△ P O B$ , it can be noticed that $P O = P O$ by using the reflexive property.

That implies, $P O ≅ P O$ .

Therefore, it can be seen that $A O = O B$ , $P O = P O$ and $P A = P B$ .

Therefore, the triangles $△ P O A$ and $△ P O B$ 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 $\bar{A B}$	Given
$\bar{A O} ≅ \bar{B O}$	Definition of midpoint
$\bar{P A} ≅ \bar{P B}$	Given
$\bar{P O} ≅ \bar{P O}$	Reflexive property
$△ P O A ≅ △ P O B$	SSS postulate

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Write proofs in two-column form.
Given: Plane $M$ bisects $\bar{A B}$ ; $\bar{P A} ≅ \bar{P B}$
Prove: $△ P O A ≅ △ P O B$

Short Answer

Step by step solution

Step 1. Observe the figure.

Step 2. Description of step.

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

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Write proofs in two-column form.Given: PlaneM bisects AB¯;PA¯≅PB¯Prove:△POA≅△POB

Short Answer

Step by step solution

Step 1. Observe the figure.

Step 2. Description of step.

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

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Calculus

Discrete Mathematics

Pure Maths

Geometry

Decision Maths

Logic and Functions

Study anywhere. Anytime. Across all devices.

Write proofs in two-column form.
Given: Plane $M$ bisects $\bar{A B}$ ; $\bar{P A} ≅ \bar{P B}$
Prove: $△ P O A ≅ △ P O B$