Question

prove that the sum of all angles of a triangle is 180°.

Answer 1

To prove: <A + <B + <C = 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 through the point parallel to the line.

<A = <ABD

[ Alternate interior angles cut by Transversal AB cutting parallel lines AC and DE must be equal.]

<C = <CBE

[ alternate interior angles formed by Transversal CB lines AC and DE must be equal.]

<ABD + <B + <CBE = 180°

[ Their sum is a straight angle.]

<A + <B + <C = 180°

[ Equals may be substituted for equals.]

Hope it helps u

@skb

Answer 2

$Let \: \: PQR \: \: be \: \: a \: \: triangle.$

$\frac{To \: \: Prove : - \: ∠P \: + \:∠Q \: + \: ∠R = 180°}{}$

$\frac{Const : - Through \: \: P \: \: draw \: \: CD ||QR}{}$

$\frac{Proof : - }{}$

$Since \: \: CD || QR \: \: and \: \: PQ \: \: is \: \: the \: \: transversal. \\ \\ ∠ 1 = ∠ 4 \: \: (alternate \: \: angles)$

$Again ,\: \: CD || QR \: \: and \: \: PR \: \: is \: \: the \: \: transversal. \\ \\ ∠ 3 = ∠ 5 \: \: \: (alternate \: \: angles)$

$Adding \: ∠2 \: on \: \: the \: \: both \: \: sides \\ \\ ∠ 1 + ∠2 + ∠ 3 = ∠4 + \:∠2+ \: ∠ 5$

$= >∠ 4 +∠ 2+∠ 5 = 180° \: (linear \: \: pair) \\ \\ = > ∠1 + ∠2 +∠ 3 = 180° \\ \\ = >∠ P +∠Q +∠ R = 180° \: \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \frac{Proved}{ \frac{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: }{} }$