Q. 24.3( 23 Votes )

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°.

Answer :


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)


Add (1), (2) and (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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