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Given: Quadrilateral DEFG

is inscribed in circle P.

Prove: m∠D+m∠F=180∘


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

Given: Quadrilateral DEFG is inscribed in circle P. Prove: m∠D+m∠F=180∘ Drag and drop-example-1
Given: Quadrilateral DEFG is inscribed in circle P. Prove: m∠D+m∠F=180∘ Drag and drop-example-1
Given: Quadrilateral DEFG is inscribed in circle P. Prove: m∠D+m∠F=180∘ Drag and drop-example-2
User XKxAxKx
by
7.1k points

2 Answers

2 votes

Answer:

The answer is down below.

Explanation:

I took the quiz. I hope this helps :)

Given: Quadrilateral DEFG is inscribed in circle P. Prove: m∠D+m∠F=180∘ Drag and drop-example-1
User Nikita Khandelwal
by
5.9k points
5 votes
we know that
The inscribed angle Theorem states that the inscribed angle measures half of the arc it comprises.
so
m∠D=(1/2)*[arc EFG]
and
m∠F=(1/2)*[arc GDE]

arc EFG+arc GDE=360°-------> full circle

applying multiplication property of equality
(1/2)*arc EFG+(1/2)*arc GDE=180°

applying substitution property of equality
m∠D=(1/2)*[arc EFG]
m∠F=(1/2)*[arc GDE]
(1/2)*arc EFG+(1/2)*arc GDE=180°----> m∠D+m∠F=180°

the answer in the attached figure
Given: Quadrilateral DEFG is inscribed in circle P. Prove: m∠D+m∠F=180∘ Drag and drop-example-1
User Pelotasplus
by
7.1k points
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