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Class 9
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Math
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Sides And Angles Of A Triangle
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Theorem 2

Sides And Angles Of A Triangle

Theorem 2

Math

Prove that: Prove that if two angles of a triangle are unequal in measure, the side opposite to the greater angle is longer than the side opposite to the smaller angle.
Given: In △ABC,\triangle ABC,△ABC, m∠Bm\angle Bm∠B > m∠Cm\angle Cm∠C
To prove: mAC‾m\overline{AC}mAC>AB‾\overline{AB}AB,
Construction: Make ∠ABM≅∠C.\angle ABM \cong \angle C.∠ABM≅∠C. Draw BN‾\overline{BN}BN, the bisector of ∠MBC\angle MBC∠MBC, i.e. m∠1=m∠2m\angle 1=m\angle 2m∠1=m∠2.

Explanation

  • To prove this theorem statement, we start by assuming that we have a triangle ABC with angle B greater than angle C.

  • Then, we draw a line from point B to a point M on side AC, such that angle ABM is congruent to angle C. This creates two new angles, ABN and MBN, which we can use to prove our statement.

  • We start by showing that angle ABN is congruent to angle ANB. This is because angle ANB is the sum of angles C and 2, while angle ABN is the sum of angles ABM and 1.

  • We know that angles ABM and C are congruent because they are alternate interior angles formed by the transversal BM and the parallel lines AB and AC.

  • We also know that angles 1 and 2 are congruent because they are both bisectors of angle MBC. Therefore, angles ABN and ANB are congruent by the angle addition postulate.

  • Next, we use this congruence to show that side AC is longer than side AB. This is because if two angles of a triangle are congruent, then the sides opposite to those angles are congruent as well.

  • In other words, AB is congruent to AN. Therefore, since angle ABN is congruent to angle ANB, we know that AB is congruent to AN. But AN is a segment of side AC, so we can conclude that AC is longer than AB.