D AY 81 β P ROPORTIONAL TRIANGLES

I NTRODUCTION In previous lessons, we have learned that a line parallel to one side of a triangle divides the other sides in equal proportions. In this lesson, we will use this idea to show how drawing such a line results to a triangle which is similar to the original triangle.

V OCABULARY Scalar This is any real number and not a vector or a matrix.

Drawing a line parallel to one side of a triangle results to a triangle which is similar to the original triangle. Consider the βπππ with a line AB parallel to side MN below. O A B N π

We will show that βπ΅πΆπ·~βπππ. β πππ = β π΅ππΆ. Since β π is shared. Now we show corresponding pairs of sides which include these angles are proportional. ππ ππ That is, ππ΅ = ππΆ Since π΅πΆ β₯ ππ , then from triangle proportionality theorem, AB divides sides MO and NO proportionally. π΅π πΆπ Therefore, ππ΅ = ππΆ β¦ β¦ (π) Adding 1 to both sides of equation (π) we get, π΅π πΆπ ππ΅ + 1 = ππΆ + 1 β¦ β¦ (ππ) Equation (ππ) can be written as, π΅π ππ΅ πΆπ ππΆ ππΆ which can be written as below ππ΅ + ππ΅ = ππΆ +

π΅π+ππ΅ πΆπ+ππΆ = β¦ β¦ πππ ππ΅ ππΆ Since ππ΅ + π΅π = ππ and ππΆ + πΆπ = ππ , we substitute these equations in equation (πππ) to get, ππ ππ ππ΅ = ππΆ Thus, the corresponding pairs of sides of the two triangles are proportional. Since the corresponding pairs of angles are equal and the corresponding pairs of sides are proportional then by S.A.S criterion βπ΅πΆπ·~βπππ. This means that we multiply the length of each side of βπππ by the same scalar to get the length of the corresponding side of βπ΅ππΆ.

Example Find the value of π in the triangle below. 10 ππ 8 ππ π 10 ππ

Solution π+10 10 10 = 8 8 π + 10 = 10 Γ 10 8π + 80 = 100 8π = 20 π = 2.5 ππ

HOMEWORK Find the value of π in the diagram below 3 ππ π 9 ππ 8 ππ

A NSWERS TO HOMEWORK 2 ππ

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