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Medians and Altitudes of a Triangle Geometry. 3/19/2016 2 Objectives: Use properties of medians of a triangle Use properties of altitudes of a triangle.

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Presentation on theme: "Medians and Altitudes of a Triangle Geometry. 3/19/2016 2 Objectives: Use properties of medians of a triangle Use properties of altitudes of a triangle."— Presentation transcript:

1 Medians and Altitudes of a Triangle Geometry

2 3/19/2016 2 Objectives: Use properties of medians of a triangle Use properties of altitudes of a triangle

3 3/19/2016 3 Medians of a triangle A median of a triangle is a segments whose endpoints are a vertex of the triangle and the midpoint of the opposite side. For instance in ∆ABC, shown at the right, D is the midpoint of side BC. So, AD is a median of the triangle

4 3/19/2016 4 Centroids of the Triangle The three medians of a triangle are concurrent (they meet). The point of concurrency is called the CENTROID OF THE TRIANGLE. The centroid, labeled P in the diagrams in the next few slides are ALWAYS inside the triangle.

5 3/19/2016 5 CENTROIDS - ALWAYS INSIDE THE TRIANGLE

6 3/19/2016 6 THEOREM about Concurrency of Medians of a Triangle The medians of a triangle intersect at a point that is two thirds of the distance from each vertex to the midpoint of the opposite side. If P is the centroid of ∆ABC, then AP = 2/3 AD, BP = 2/3 BF, and CP = 2/3 CE

7 3/19/2016 7 So what? The centroid of a triangle can be used as its balancing point. Let’s try it. I’ve handed out triangle to each and every one of you. Construct the medians of the triangles in order to great the centroid in the middle. Then use your pencil to balance your triangle. If it doesn’t balance, you didn’t construct it correctly.

8 3/19/2016 8 Ex. 1: Using the Centroid of a Triangle P is the centroid of ∆QRS shown below and PT = 5. Find RT and RP.

9 3/19/2016 9 Ex. 1: Using the Centroid of a Triangle Because P is the centroid. RP = 2/3 RT. Then PT= RT – RP = 1/3 RT. Substituting 5 for PT, 5 = 1/3 RT, so RT = 15. Then RP = 2/3 RT = 2/3 (15) = 10 ► So, RP = 10, and RT = 15.

10 3/19/2016 10 Ex. 2: Finding the Centroid of a Triangle Find the coordinates of the centroid of ∆JKL You know that the centroid is two thirds of the distance from each vertex to the midpoint of the opposite side. Choose the median KN. Find the coordinates of N, the midpoint of JL.

11 3/19/2016 11 Ex. 2: Finding the Centroid of a Triangle The coordinates of N are: 3+7, 6+10 = 10, 16 2 2 2 2 Or (5, 8) Find the distance from vertex K to midpoint N. The distance from K(5, 2) to N (5, 8) is 8-2 or 6 units.

12 3/19/2016 12 Ex. 2: Finding the Centroid of a Triangle Determine the coordinates of the centroid, which is 2/3 ∙ 6 or 4 units up from vertex K along median KN. ►The coordinates of centroid P are (5, 2+4), or (5, 6).

13 3/19/2016 13 Using altitudes of a triangle An altitude of a triangle is the perpendicular segment from the vertex to the opposite side or to the line that contains the opposite side.

14 Using altitudes of a triangle An altitude can lie inside, on, or outside the triangle. Every triangle has 3 altitudes. The lines containing the altitudes are concurrent and intersect at a point called the orthocenter of the triangle. 3/19/2016 14

15 3/19/2016 15 Ex. 3: Drawing Altitudes and Orthocenters Where is the orthocenter located in each type of triangle? a.Acute triangle b.Right triangle c.Obtuse triangle

16 3/19/2016 16 Acute Triangle - Orthocenter ∆ABC is an acute triangle. The three altitudes intersect at G, a point INSIDE the triangle.

17 3/19/2016 17 Right Triangle - Orthocenter ∆KLM is a right triangle. The two legs, LM and KM, are also altitudes. They intersect at the triangle’s right angle. This implies that the ortho center is ON the triangle at M, the vertex of the right angle of the triangle.

18 3/19/2016 18 Obtuse Triangle - Orthocenter ∆YPR is an obtuse triangle. The three lines that contain the altitudes intersect at W, a point that is OUTSIDE the triangle.

19 3/19/2016 19 Theorem about Concurrency of Altitudes of a triangle The lines containing the altitudes of a triangle are concurrent. If AE, BF, and CD are altitudes of ∆ABC, then the lines AE, BF, and CD intersect at some point H.


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