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5.4 Medians and Altitudes. Vocabulary…  Concurrent- 3 or more lines, rays, or segments that intersect at the same point  Median of a Triangle – a segment.

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Presentation on theme: "5.4 Medians and Altitudes. Vocabulary…  Concurrent- 3 or more lines, rays, or segments that intersect at the same point  Median of a Triangle – a segment."— Presentation transcript:

1 5.4 Medians and Altitudes

2 Vocabulary…  Concurrent- 3 or more lines, rays, or segments that intersect at the same point  Median of a Triangle – a segment from a vertex to the midpoint of the opposite side  Centroid – point of concurrency of 3 medians  Altitude of a Triangle – the perpendicular segment from a vertex to the opposite side or to the line that contains the opposite side – there are 3 in a

3  Orthocenter - the point where the 3 altitudes of a intersect  Theorem 5.8: Concurrency of Medians of a -the medians of a intersect at a point that is 2/3 the distance from each vertex to the midpoint of the opposite side. Theorem 5.9: Concurrency of Altitudes of a -the lines containing the altitudes of a are concurrent

4 SQ = 2 3 SW Concurrency of Medians of a Triangle Theorem 8 =8 = 2 3 SW Substitute 8 for SQ. 12 = SW Multiply each side by the reciprocal,. 3 2 Then QW = SW – SQ = 12 – 8 =4. So, QW = 4 and SW = 12. In RST, Q is the centroid and SQ = 8. Find QW and SW.

5  Practice In Exercises 1–3, use the diagram. G is the centroid of ABC. 1. If BG = 9, find BF. ANSWER 13.5 2. If BD = 12, find AD. ANSWER 12 3. If CD = 27, find GC. ANSWER 18

6  Find the orthocenter. Find the orthocenter P in an acute, a right, and an obtuse triangle. (Draw 3 altitudes…drop perpendicular lines from vertex to opposite side.) SOLUTION Acute triangle P is inside triangle. Right triangle P is on triangle. Obtuse triangle P is outside triangle.

7 Can you answer these??????  Look at the and answer the following:  1. Is BD a median of ABC?  2. Is BD an altitude ABC?  3. Is BD a perpendicular bisector? B B   A D C A D C


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