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The side BC of triangle ABC is produced on both sides. Prove that the sum of the two exterior angles so formed is greater than angle A by 180 degree.

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∠ ABD + ∠ ABC = 180° ….(i) (Angle on the same line)

∠ ACE + ∠ACB = 180° ……….(ii)

Adding (i) and (ii), we get

∠ ABD + ∠ ABC + ∠ ACE + ∠ACB = 360°

∠ ABD + ∠ ACE + ( ∠ ABC + ∠ACB ) =360° …………..(iii)

Now, In ΔABC,

∠ ABC + ∠ ACB + ∠ BAC = 180°

∠ ABC + ∠ ACB = 180°- ∠ BAC

Putting the value of ∠ ABC + ∠ ACB in eq.(iii), we get

∠ ABD + ∠ ACE +180°- ∠ BAC =360°

∠ ABD + ∠ ACE = 360°- 180°+ ∠ BAC

∠ ABD + ∠ ACE = 180°+ ∠ BAC

Which proves that sum of the two exterior angles formed is greater than angle A by 180°.

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in tri< abc,

<A+<B+<C=180*_______1

<B+ext<B=180* (LINEAR PAIR)______2

<C+ext<C=180* (  "  )________3

ADDING 2 AND 3,

<B+<C+ext<B+ext<C=360*_______4

MULTIPLYING 1 WITH 2,WEGET,

2<A+2<B+2<C=360*_______5

EQUATING 4 AND 5 WE GET,

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         To Prove:               ext.angle B {b}  +ext.angle C {c}  = angle A + 180 degree

A                                   In triangle ABC,

                                                     <A+<B+<C=180....(1)       [by angle sum property]

                     <B+<b=180                             <C+<c=180      [ linear pair]

                     <B=180-<b.....(2)                    <C=180-<c......(3)

                            Putting the values of equ. (2) and (3) in equ. (1) we get:

                                                 <A+180-<b+180-<c=180

                            A-b-c=180

                                             A+180=b+c      H.P.                  

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