Question

prove that the measure of an exterior angle of triangle is equal to the sum of its remote interior angles​

Answer 1

Given:-

• A Δ ABC , BC is produced to D, where ACD is the exterior angle of Δ ABC.

To Prove:-

• ∠A + ∠B = ∠ACD

Property used:-

• Angle Sum Property of Triangle.

• Sum of Linear Angle is 180°.

Solution:-

In Δ ABC,

∠A + ∠B + ∠ACB = 180° ____ {1}

As AD is a straight line, ∠ACB and ∠ACD are Supplementary Angles So,

∠ACB + ∠ACD = 180° ____ {2}

On Comparing {1} and {2}, we get

∠A + ∠B + ∠ACB =∠ACB + ∠ACD

${\boxed{\red{∠A + ∠B =∠ACD}}}$

HENCE PROVED.

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Answer 2

in triangle ABC extend BC to form the exterior angle at C, call that x

We know angle A + Angle B + angle C = 180

But angle C + x = 180, (by supplementary angle theorem)

so angle A + angle B + angle C = x + angle C

subtract angle C from both sides

angle A + angle B = x, the exterior angle.

Therefore, the exterior angle of a triangle has measure equal to the sum of the measures of the 2 interior angles remote from it