Prove that if one side of a triangle is longer than another side, then the angle opposite the longer side is larger than the angle opposite the shorter side(Write a two-column proof)Given, BC > BA Prove m z BAC > mz CUse the paperclip button below to attach filesBIUCalcProjectStudentFi..odsOpen file77°F ClearF3F7F8F9E10F11F12Prt ScInsert#24&大3569ERDFG HKc v B N MAltCtr

For the ∆ABC, Given BC>BA. Let D be the point on BC such that BA=BD⇒∠1=∠2 (Angles opposite to equal sides are equal).On the line BC, ⇒∠BDA+∠ADC=180° (Pair of linear angles) ⇒∠1+∠ADC=180° We know that sum of all interior angles in a triangle is always 180°. For ∆ADC, ⇒∠ADC+∠ACD+∠DAC=180°⇒⇒180°-∠1+∠ACD+∠3=180°⇒∠ACD=∠1-∠3 From the diagram we can say that, ∠ACD=∠ACB=∠C So ∠C=∠1-∠3 ↦Result 1 From the diagram we can say that, ⇒∠BAC=∠BAD+∠DAC⇒∠BAC=∠1+∠3 ↦Result 2 From Result 1 and Result 2, we can say that inequality formed here is, ⇒∠1+∠3>∠1-∠3⇒∠BAC>∠C Hence proved.