Question

Prove that the
that the sum of a angle of
triangle is 180 degree​

Answer 1

Step-by-step explanation:

Prove that sum of all three angles is 180°

Given: ∆ABC

To prove: ∠A + ∠B + ∠C = 180°

Construction: Draw PQ || BC, passes through point A.

Proof: ∠1 = ∠B   and  ∠3 = ∠C         ……. (i)

[∵ alternate angles ∵ PQ || BC]

∵ PAQ is a line

∴∠1 + ∠2 + ∠3 = 180°     (linear pair)

∠B + ∠2 + ∠C = 180°

∠B + ∠CAB + ∠C = 180°

Proved.

Answer 2

$\huge\tt \underline \green{ {\: given : - }}$

$\: \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: ΔPQR \: with \: angle \: ∠1, \: ∠2 \: and \: ∠3$

$\huge \tt \underline \green{Prove : - }$

$\: \tt \: \: \: \: \: \: \: \: \: \: \: \: \: \: ∠1 + ∠2 + ∠3 = 180°$

$\huge \tt \underline \green{Construction : - }$

$\: \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: Draw \: a \: line \: XY \: passing \: through \: P \: parallel \: to \: QR.$

$\huge \tt \underline \green{Proof:-}$

$\sf \: For \: line \: XY \: and \: QR, \: with \: transversal \: PQ \\ \sf \: ∠2 = ∠4 \: \: \: \: \: \: \: \: \: \bigg( Alternate angles \bigg) \: \: ... (1)$

$\sf \: For \: line \: XY \: and \: QR, \: with \: transversal \: PQ \\ \sf \: ∠3 = ∠5 \: \: \: \: \: \: \: \: \: \bigg( Alternate angles \bigg) \: \: ... (2)$

$\sf \blue {Also, \: for \: line \: XY} \\ \sf \: For \: ∠1 + ∠4 + ∠5 =180° \: \: \: \: \bigg( Linear pair \bigg)$

$\sf \: For \: ∠1 + ∠2+ ∠3 =180° \: \: \: \: \bigg( From (1) and (2) \bigg)$

$\sf \: Hence, sun \: of \: all \: angle \: of \: a \: triangle \: are \: equal$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf\huge \: \blue {Hence \: \: proved }$