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Aim: What’s so special about these lines of a triangle?

Aim: What’s so special about these lines of a triangle?. Do Now:. l | | p Find the measure of x, y and z. l. p. altitudes meet at one point called the orthocenter. The Special Lines of a Triangle Altitude BH is an altitude from B to AC

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Aim: What’s so special about these lines of a triangle?

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  1. Aim: What’s so special about these lines of a triangle? Do Now: l | | p Find the measure of x, y and z l p Course: Applied Geometry

  2. altitudes meet at one point called the orthocenter. The Special Lines of a Triangle Altitude BH is an altitude from B to AC Altitude of a Triangle - A line segment from a vertex and perpendicular to the opposite side. Each triangle has three vertices Each triangle has three altitudes. Course: Applied Geometry

  3. The Special Lines of a Triangle Altitude BH is an altitude from B to AC Altitude of a Triangle - A line segment from a vertex and perpendicular to the opposite side. obtuse triangles- has two external altitudes and one internal. Course: Applied Geometry

  4. bisectors meet at one point called the incenter. Angle Bisector BQ is the bisector of  B: mABQ = mCBQ Angle bisector of a triangle - A line segment that divides an angle of a triangle into two halves. Each triangle has three bisectors. Course: Applied Geometry

  5. medians meet at one point called the centroid. Median BM is the median from B to the midpoint of AC: AM = MC Median of a triangle - A line segment from a vertex of a triangle to the midpoint of the opposite side. Each triangle has three medians. Course: Applied Geometry

  6. Median BM is the median from B to the midpoint of AC AM = MC Median of a triangle - A line segment from a vertex of a triangle to the midpoint of the opposite side. Right Triangle Equilateral Triangle Special lines of various triangles Course: Applied Geometry

  7. BM  MC A In the same ABC, BM = 3x – 2 and CM = x + 8. Find the value of x and BC N C B M A In ABC, AM is median to BC. If AB = 16, AC = 7, and BC = 11, find BM. N C = 5.5 B M BM + MC = BC = 11 3x – 2 = x + 8 2x – 2 = 8 2x = 10 x = 5 BC = BM + CM = (3x – 2) + (x + 8) BC = BM + CM = (3·5 – 2) + (5 + 8) BC = BM + CM = (13) + (13) = 26 Course: Applied Geometry

  8. Aim: What’s so special about these lines of a triangle? Do Now: In ABC, AD is an altitude to BC. If ADB = 3x – 30, find the value of x. A B D C ADB = 900 = 3x - 30 120 = 3x 40 = x Course: Applied Geometry

  9. In SRT, RP is an angle bisector. If SRP = 2x + 10 and mSRT = 84, find the value of x. SRP R T S P 840 In ABC, AD is an altitude to BC. If ADB = 3x – 30, find the value of x. A B D C ADB = 900 = 3x - 30 120 = 3x 40 = x SRP = 1/2 (mSRT) = 1/2 · 84 = 42 2x + 10 = 42 2x = 32 x = 16 Course: Applied Geometry

  10. 700 300 200 B D C + mBAD + 90 20 = 180 A In ABC, AD is an altitude to BC. 1000 If ABD = 20 and BAC = 100, find ACB. 200 C B D Sum of s of  = 1800 mABD + mBAD + mADB = 180 mBAD = 70 Next? If mBAD = 70, then mDAC = 30 A 900 mADC + mCAD + mACD = 180 90 + 30 + mACD = 180 Course: Applied Geometry mACD = 60

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