16+Triangle+Proportionality+Theorems

Triangle Proportionality Theorem

If a line parallel to one side of a triangle intersects the other two sides of the triangle, then the line divides these two sides proportionally. In the given triangle ABC, BC is the base of the triangle. DE is drawn parallel to BC and it intersects the other two sides AB and AC at D and E respectively. So, AD/DB = AE/EC

Ex)

Find the value of x. The lines //QR// and //ST// are parallel.

Therefore, by the Triangle Proportionality Theorem,

//PS/QS// = //PT/RT//

Substitute the values and solve for x.

//6/2 = 9/x//

Cross multiply.

6x = 18

Divide both sides by 6.

//6x/6 = 18/6//

//x = 3//

The value of x is 3.

Consider a triangle PQR, where QR is the base of the triangle. If a line ST is drawn parallel to the base (QR), therefore DE || BC. The values are PS =4, PT =4, TR =!2, Find the value of SQ. [(PS)/(PQ)] [(PT)/(PR)] [Proportionality Theorem]

[PS/(PS+SQ)] = [PT/(PT+PR)]

Substituting the values,

[4/(x+4)] [4/(4+12)]

On Cross-multiplying, we get,

16 + 48 = 4(x) +16,

64 = 4(x) +16,

4(x) = 64-16

4(x) = 48

Dividing by 4 on both sides,

[4(x)/4] = [48/4]

On simplifying we get,

x = 12



helpful links http://www.channels.com/episodes/show/3546709/Triangle-Proportionality-and-Triangle-Angle-Bisector-Theorem http://www.icoachmath.com/sitemap/Triangle_Proportionality_Theorem.html http://hotmath.com/hotmath_help/topics/triangle-proportionality-theorem.html http://www.malinc.se/math/geometry/transversalen.php