Given triangle ABC with exterior angle <BCD, Prove m<A + m<B= m<BCD

Given: triangle ABC with exterior angle <BCD

Prove: m<A + m<B= m<BCD

.........B
..........•
......../...\
......./.....\
....../.......\
...../.........\
..../ ..........\
A •------------•--------•
....................C.........D
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1 Answer

Interior angle C=180-(A+B) (angles of triangle add up to 180 degrees) and exterior angle C (BCD) = 180-BCA (complementary angles on a straight line) =180-(180-(A+B))=A+B.

by Top Rated User (1.1m points)

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