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Drag and drop an answer to each box to correctly complete the proof.

Given: m∥nm∥n , m∠1=50∘m∠1=50∘ , m∠2=48∘m∠2=48∘ , and line s bisects ∠ABC∠ABC .

Prove: m∠3=49∘

It is given that m∥nm∥n , m∠1=50∘m∠1=50∘ , m∠2=48∘m∠2=48∘ , and line s bisects ∠ABC∠ABC . By the ,  m∠DEF=98∘m∠DEF=98∘ . Because  angles formed by two parallel lines and a transversal are congruent,  ∠DEF≅∠ABC∠DEF≅∠ABC , so m∠ABC=98∘m∠ABC=98∘ . By the , angles 4 and 5 are congruent, and m∠4m∠4 is half  m∠ABCm∠ABC . So the measure of  m∠4=49∘m∠4=49∘ . Because vertical angles are congruent, ∠3≅∠4∠3≅∠4 . Finally, m∠3=m∠4m∠3=m∠4 by the angle congruence postulate, so  m∠3=49∘m∠3=49∘ by the .

 

 

 

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 Oct 18, 2018
 #1
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Here is a proof

 

Angle DEF  =  angle ABC              

A transversal cutting two parallel lines makes alternate exterior angles equal

 

Sum of  measures of angles 1 and 2  = 50° + 48°  =  98°   = measure of angle DEF  =  measure of angle ABC

 

But since  s bisects   angle ABC, then angle 4  =  98 / 2   = 49°

 

But angles 3 and 4  are vertical angles....therefore.... measure of angle 3  =  49°

 

 

cool cool cool

 Oct 18, 2018

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