Prove that the sum of exterior angles of a triangle obtained by extending its sides in same direction is 360°
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Prove that the sum of exterior angles of a triangle obtained by extending its sides in same direction is 360°.
Given
- ∠PAN , ∠AND and ∠ACT are exterior angles of triangle ABC
To prove
∠PAB + ∠QBC + ∠ACR = 360°
Proof
Considering exterior ∠PAB OF triangle ABC,
∠ABC and ∠ACB are the it's remote interior angles
∠PAB = ∠ABC + ∠ACB ----(I)
Similarly,
∠ACR = ∠ABC + ∠BAC ----(II)..theorem of remote interior angles and ∠CBQ = ∠BAC + ∠ACB----(III)
Adding (I) , (II) and (III)
∠PAB + ∠ACR + ∠CBQ
= ∠ABC + ∠ACB + ∠ABC + ∠BAC + ∠ACB
= 2∠ABC + 2∠ACB + 2∠BAC
= 2 ( ∠ ABC + ∠ ACB + ∠BAC )
= 2 × 180°...Sum of interior angles of a triangle
= 360°
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