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Given: <1 and <2 are supplementary, and <2 and <3 are supplementary. Prove: <1 is congruent to <3

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Explanation:

If ∠1 and ∠2 are supplementary, then m∠1 + m∠2 = 180° (1).

If ∠2 and ∠3 are supplementary, then m∠2 + m∠3 = 180° (2).

From (1) we have

m∠1 + m∠2 = 180° /subtract m∠1 from both sides

m∠2 = 180° - m∠1

substitute it to (2)

(180° - m∠1) + m∠3 = 180°

180° - m∠1 + m∠3 = 180° /subtract 180° from both sides

- m∠1 + m∠3 = 0 /add m∠1 to both sides

m∠3 = m∠1 ⇒ m∠1 = m∠3 ⇒ ∠1 ≅ ∠3

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