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 Q. To verify that the sum of the interior angles of a triangle is 180.

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To prove:  ÐA + ÐB + ÐC = 180°
 Draw line through B parallel to AC    Given a line and a pt. not on and label two points D and E, one     the line, exactly one and only on each side of B                     one line can be drawn thru                                       the point parallel to the line.           drawing%28400%2C170%2C-3%2C3%2C-1%2C2%2C+%0D%0Atriangle%28-1%2C0%2C2%2C0%2C0%2C1.5%29%2C%0D%0Alocate%28-1%2C0%2CA%29%2C%0D%0Alocate%282%2C0%2CC%29%2C%0D%0Alocate%280%2C1.8%2CB%29%2C%0D%0Aline%28-2%2C1.5%2C2%2C1.5%29%2Clocate%28-2%2C1.5%2CD%29%2Clocate%282%2C1.5%2CE%29%0D%0A+%29  ÐA = ÐABD                             Alternate interior angles formed by                                       Transversal AB cutting parallel                                        lines AC and DE must be equal.  ÐC = ÐCBE                               alternate interior angles formed by                                       Transversal CB cutting parallel                                        lines AC and DE must be equal.  ÐABD + ÐB + ÐCBE = 180°               Their sum is a straight angle.  ÐA + ÐB + ÐC = 180°                   Equals may be substituted for equals.  Thumbs up please

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 Hi!

Here is the answer to your query.
 
Consider a triangle PQR and ∠1, ∠2 and ∠3 are the angles of ΔPQR (figure shown below). We need to prove that ∠1 + ∠2 + ∠3 = 180°.
 
XPY is a line.
∴∠4 + ∠1 + ∠5 = 180°  … (1)
But XPY || QR and PQ, PR are transversals.
So, ∠4 = ∠2 and ∠5 = ∠3  (Pairs of alternate angles)
Substituting ∠4 and ∠5 in (1), we get
∠2 + ∠1 + ∠3 = 180°
∴∠1 + ∠2 + ∠3 = 180°
Thumsb up please

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