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2.3: Angle Properties in Triangles Can you prove that the sum of the measures of the interior angles of any triangle is 180o? A. Draw an acute triangle, ∆RED. Construct line PQ through vertex D, parallel to RE. B. Identify pairs of equal angles in your diagram. Explain how you know that the measures of the angles in each pair are equal. C. What is the sum of the measures of ∠PDR, ∠RDE, and ∠QDE? Explain how you know. D. PDR =DRE; QDE = RED Alternate interior angles 180°, because the three angles form a straight line Explain why ∠𝑫𝑹𝑬 + ∠𝑹𝑫𝑬 + ∠𝑹𝑬𝑫 = 𝟏𝟖𝟎𝒐 . (prove) 1. 2. 3. 4. E. Statement PDR +∠RDE +∠QDE = 180° ∠DRE = ∠PDR ∠RED = ∠QDE DRE +∠RDE +∠RED = 180° Reason These angles form a straight line Alternate interior angles Alternate interior angles Substitute angles in step 2 & 3 in equation in step 1. Repeat parts A to D, first for an obtuse triangle and then for a right triangle. Are your results the same as they were for the acute triangle? Example 1 – In the diagram, ∠MTH is an exterior angle of ∆MAT. Determine the measures of the unknown angles in ∆MAT. MTA + MTH = 180o MTA + 155o = 180o MTA = 180o – 155o = 25o TAM + AMT + MTA = 180o TAM + 40o + 25o = 180o TAM = 180o – 40o -25o = 115o If you are given one interior angle and one exterior angle of a triangle, can you always determine the other interior angles of the triangle? Explain, using diagrams. If the interior angle you are given is adjacent to a known exterior angle, then you cannot determine the other angles. For example, in ∆PQR, neither of the non-adjacent interior angles are known, so there is not enough information to determine the unknown interior angles. Example 2 – Determine the relationship between an exterior angle of a triangle and its non-adjacent interior angles. d + c = 180o on the same line c + a + b = 180o angles in a triangle add to 180o d + c = c + a + b since the 2 equations above both equal 180o they must also equal each other d = a + b get d by itself So an exterior angle is equal to the sum of the two non-adjacent interior angles of a triangle. Example 3 – (a) Determine the measures of ∠NMO, ∠MNO, ∠QMO. LMN = MNP = 67o MNP = MNO + PNO (Alt int) 67o = MNO + 20o MNO = 67o – 20o = 47o NMO + MON + ONM = 180o (SATS) NMO + 39o + 47o = 180o NMO = 180o – 39o – 47o = 94o MNP + NMQ = 180o (int. ’s on same side of trans. suppl.) 67o + NMQ = 180o NMQ = 180o – 67o =113o NMQ = NMO + OMQ 113o = 94o + OMQ OMQ = 113o – 94o = 19o (b) If QP∥MR, determine the measures of ∠MQO, ∠MOQ, ∠NOP, ∠OPN, and ∠RNP. LMN = MQO (Corr. ’s) = 67o MOQ + OQM + QMO = 180o MOQ + 67o + 19o = 180o MOQ = 180o – 67o – 19o =94o MNP + OPN = 180o (int ’s same side) 67o + OPN = 180o OPN = 180o – 67o = 113o OPN + PNO + NOP = 180o (SATS) 113o + 20o + NOP = 180o NOP = 180o – 113o – 20o = 47o RNP = NPO (Alt Int) = 113o Example 4: Prove: A = 30°v ∆BCD is an equilateral triangle which means all 3 sides are the same and all 3 angles are the same. If all of the angles need to add to 180o and all of the angles are the same, the angles must each equal 180o 3 So, CBD = BCD = CDB = 180o 3= 60o ∆ADB is an isosceles triangle. This means that 2 sides are equal (AD = BD) and 2 angles are equal (DBA = BAD). BDA = 180o - CDB = 180o – 60o = 120o This means the other two angles in ∆ADB need to add to 60o (because all three need to add to 180o and one of the angles is known to be 120o). We also know that the two angles we have left need to have the same measure (since this is an isosceles triangle). DBA = BAD = 60o 2 = 30o Therefore: A = 30o Key Ideas: You can prove properties of angles in triangles using other properties that have already been proven. In any triangle, the sum of the measures of the interior angles is proven to be 180o. The measure of any exterior angle of a triangle is proven to be equal to the sum of the measures of the two non-adjacent interior angles. 2.3 Assignment: Nelson Foundations of Mathematics 11, Sec 2.3, pg. 90-93 Questions: 2, 3, 6, 7, 9, 12a, 14