Question

The side BC of triangle ABC is produced to form ray BD,CE is drawn parallel to AB. Show that angle ACD is equal to angle A angle B. Prove that angleA + angleB + angleACB=180°

Answer 1

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Answer 2

Answer:

Given: Δ ABC is given.

        BC produced to form BC ray and CE is parallel to AB

To prove: ∠A + ∠B + ∠ACB = 180°

Proof,

∠A + ∠B = ∠ACD  

∠ ACB + ∠ACD = 180° ( Linear Pair )

⇒ ∠ ACD = 180° - ∠ACB

Put value of ∠ ACD

∠A + ∠B =  180° - ∠ACB

∠A + ∠B + ∠ACB =  180°

Hence Proved, Sum of all angles of a triangle is 180°.