# Sides BC, CA and AB of a triangle ABC are produced in an order, forming exterior angles ∠ACD, ∠BAE, and ∠CBF. Show that ∠ACD + ∠BAE + ∠CBF = 360°.

As we know, using theorem (2) i.e. if a side of the triangle is produced, the exterior angle so formed is equal to the sum of corresponding opposite interior angles.

So,

ACD = BAC + ABC …(1)

BAE = ABC + ACB …(2)

CBF = BAC + BCA …(3)

We get,

ACD+ BAE+ CBF = BAC+ ABC+ ABC + BCA+ BAC + BCA

⇒∠ACD+BAE+CBF = 2( BAC+ ABC BCA)

As we know, using theorem (1), in any triangle, sum of the three interior angles is 180°.

So,

ABC+ BAC + BCA = 180°

ACD+BAE+CBF = 2(180°)

ACD+BAE+CBF = 360°

Hence proved.

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