Question

A triangle ABC has interior angles x, y, and z. Prove that x+ y+ z= 180

Answer 1

$\underline { \blue { Given : }}$

ABC is a triangle.

$\angle A = x , \angle B = y ,\: and \: \angle C = z$

$\underline { \blue { To \:prove : }}$

$x + y + z = 180\degree$

$\underline { \blue { Construction : }}$

Produce BC to a point D and through C draw a line CE parallel to BA.

$\underline { \blue { Proof : }}$

$BA \:parallel \:to\: CE \: [ By \: construction ]$

$\angle {ABC} = \angle {ECD}\: ---(1)\\ ( By \: corresponding \: angles \:axiom )$

$\angle {BAC} = \angle {ACE}\: ---(2)\\(Alternate\:interior \: angles \:for \:the \: parallel\\lines \:AB \: and \:CE )$

$\angle {ACB} = \angle {ACB}\:---(3) \\ [ Same \:angle ]$

/* Adding above three equations , we get */

$\angle {ABC} + \angle {BAC}+\angle {ACB}\\ = \angle {ECD} + \angle {ACE}+\angle {ACB}$

$But \: \angle {ECD} + \angle {ACE}+\angle {ACB} = 180\degree \\(Sum \: of \:angles \: at \: a \: point \: on \: a \: straight \: line )$

Therefore.,

$\angle {ABC} + \angle {BAC}+\angle {ACB}\\ = 180\degree$

$\implies \pink { x + y + z = 180\degree }$

$Hence , Proved.$

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