DRILL If M is the midpoint of AB and MA = 7, find AB and MB.

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Presentation transcript:

DRILL If M is the midpoint of AB and MA = 7, find AB and MB. 2) What is the mean(average) of 17 and 27? 3)Solve for x: 4x - 10 = 2x + 30 AB = 14 & MB = 7 Mean is 22 X = 20

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Midsegment of a Triangle * A midsegment of a triangle is a segment that connects the midpoints of two sides of a triangle. * The Midsegment of a Triangle is equal to one-half the third side, as well as parallel to the third side.

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Examples x =10 3x 4x + 20

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Ex. 1: Using midsegments Show that the midsegment MN is parallel to side JK and is half as long. Use the midpoint formula.

Solution: M= -2+6 , 3+(-1) 2 2 M = (2, 1) And N = 4+6 , 5+(-1) 2 2 M = (2, 1) And N = 4+6 , 5+(-1) 2 2 N = (5, 2) Next find the slopes of JK and MN. Slope of JK = 5 – 3 = 2 = 1 4-(-2) 6 3 Slope of MN= 2 – 1 = 1 5 – 2 3 ►Because their slopes are equal, JK and MN are parallel. You can use the Distance Formula to show that MN = √10 and JK = √40 = 2√10. So MN is half as long as JK.

Ex. 2: Using the Midsegment Theorem UW and VW are midsegments of ∆RST. Find UW and RT. SOLUTION: UW = ½(RS) = ½ (12) = 6 RT = 2(VW) = 2(8) = 16

Coordinate Proof Write a coordinate proof of the Midsegment Theorem. Place points A, B, and C in convenient locations in a coordinate plane, as shown. Use the Midpoint formula to find the coordinate of midpoints D and E.

Coordinate Proof D = 2a + 0 , 2b + 0 = a, b 2 2 2 2 E = 2a + 2c , 2b + 0 = a+c, b 2 2 Find the slope of midsegment DE. Points D and E have the same y-coordinates, so the slope of DE is 0. ►AB also has a slope of 0, so the slopes are equal and DE and AB are parallel.

Now what? ►The length of DE is half the length of AB. Calculate the lengths of DE and AB. The segments are both horizontal, so their lengths are given by the absolute values of the differences of their x-coordinates. AB = |2c – 0| = 2c DE = |a + c – a | = c ►The length of DE is half the length of AB.

Perimeter of Midsegment Triangle DF = ½ AB = ½ (10) = 5 EF = ½ AC = ½ (10) = 5 ED = ½ BC=½ (14.2)= 7.1 ►The perimeter of ∆DEF is 5 + 5 + 7.1, or 17.1. The perimeter of ∆ABC is 10 + 10 + 14.2, or 34.2, so the perimeter of the triangle formed by the midsegments is half the perimeter of the original triangle.