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Published byΆριστόδημος Παυλόπουλος Modified over 6 years ago
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A segment that connects a vertex of a triangle
to the midpoint of the opposite side. The point of concurrency of the medians of a triangle. The perpendicular segment from a vertex of a triangle to the opposite side. The point of concurrency of the altitudes of a triangle.
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AE BF CD Also, EP = 1/3AE, FP = 1/3BF, and DP = 1/3CD Therefore, AP = 2EP, BP = 2FP, and CP = 2DP
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GM 2/3 /3 6 9 3/2 GM
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FM = 2/3FK 10 = 2/3FK 15 = FK MK = 1/2FM MK = 1/2(10) MK = 5
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(4, 3) two thirds 2/ 2 4, 4 If you don't have a horizontal or vertical line as a median you would have to use the distance formula to find the length of the median, then take 2/3 of the length to find the centroid.
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concurrent
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Orthocenter Orthocenter
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Point K should be moved to (4, 9)
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∠ADB ∠CDB Altitude HL Congruence Theorem CD Midpoint
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AB ≅ CB Given ∠ADB ≅ ∠CDB All right ∠'s are ≅ BD ≅ BD Reflexive Property ∆ABD ≅ ∆CBD HL ≅ Thm ∠ABD ≅ ∠CBD CPCTC BD is an ∠ bisector Definition of bisector Statements Reasons
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