Question

prove that the sum of a triangle is 180​

Answer 1

Step-by-step explanation:

we can prove this by using alternate angle theorem and linear pair axiom

Answer 2

Solution:

$\underline{\bf Given:}\\ \\ \implies \sf ABC\;is\;a\;triangle.\\ \\ \\ \underline{\bf To\;Prove:}\\ \\ \implies \sf Sum\;of\;all\;angles\;of\;triangle,\;i.e\; \angle A+\angle B+\angle C=180^{\circ}\\ \\ \\ \underline{\bf Construction:}\\ \\ \implies \sf Draw\;a\;line\;PQ\;parallel\;to\;BC.\\ \\ \\ \underline{\bf Proof:} \\ \\ \sf Since\;PQ||BC\;and\;AC\;is\;transversal,\\ \\ \implies \sf \angle QAC = \angle ACB\;\;\;[Alternate\;Interior\;Angles]\;\;\;\;\;........(1)$

$\\ \\ \sf Since\;PQ||BC\;and\;AB\;is\;transversal,\\ \\ \implies \sf \angle PAB=\angle ABC\;\;\;\;[Alternate\;Interior\;Angle]\;\;\;\;.........(2)\\ \\ \sf Now, \angle QAB + \angle PAB = 180^{\circ}\\ \\ \implies \sf \angle QAC+\angle BAC + \angle PAB=180^{\circ}\;\;\;\;......(3) \\ \\ \sf Now,\;we\;just\;prove\;that,\;\angle QAC = \angle ACB\;and\;\angle PAB = \angle ABC.\\ \\ \sf Now,\;put\;the\;value\;in\;equation\;(3),\;we\;get,\\ \\ \implies \sf \angle ACB + \angle BAC + \angle ABC = 180^{\circ}$

$\implies \sf \angle C + \angle B + \angle A = 180^{\circ}\\ \\ \implies {\boxed{\bf \angle A + \angle B + \angle C = 180^{\circ}}}\\ \\ \bf ...........Hence\;Proved!!!..........$