CK-12 Geometry Concepts

(Elliott) #1

7.8. Triangle Proportionality http://www.ck12.org



  1. Draw 4 ABC. Label the vertices.

  2. DrawXYso thatXis onABandYis onBC.XandYcan beanywhereon these sides.

  3. Is 4 X BY∼4ABC? Why or why not? MeasureAX,X B,BY,andY C. Then set up the ratiosAXX BandY CY B. Are
    they equal?

  4. Draw a second triangle, 4 DEF. Label the vertices.

  5. DrawXYso thatXis onDEandYis onEFANDXY||DF.

  6. Is 4 X EY∼4DEF? Why or why not? MeasureDX,X E,EY,andY F. Then set up the ratiosDXX EandY EFY. Are
    they equal?


From this investigation, it is clear that if the line segments are parallel, thenXYdivides the sides proportionally.


Triangle Proportionality Theorem:If a line parallel to one side of a triangle intersects the other two sides, then it
divides those sides proportionally.


Triangle Proportionality Theorem Converse: If a line divides two sides of a triangle proportionally, then it is
parallel to the third side.


Proof of the Triangle Proportionality Theorem:


Given: 4 ABCwithDE||AC


Prove:ADDB=CEEB


TABLE7.2:


Statement Reason
1.DE||AC Given

2.^61 ∼=^62 ,^63 ∼=^64 Corresponding Angles Postulate
3. 4 ABC∼4DBE AA Similarity Postulate
4.AD+DB=AB,EC+EB=BC Segment Addition Postulate
5.ABBD=BCBE Corresponding sides in similar triangles are propor-
tional
6.ADBD+DB=ECBE+EB Substitution PoE
7.ADBD+DBDB=ECBE+BEBE Separate the fractions
8.ADBD+ 1 =ECBE+ 1 Substitution PoE (something over itself always equals



  1. 9.ADBD=ECBE Subtraction PoE


Example A


A triangle with its midsegment is drawn below. What is the ratio that the midsegment divides the sides into?


The midsegment’s endpoints are the midpoints of the two sides it connects. The midpoints split the sides evenly.
Therefore, the ratio would bea:aorb:b. Both of these reduce to 1:1.


Example B


In the diagram below,EB||CD. FindBC.


Use the Triangle Proportionality Theorem.


10


15


=


BC


12


−→ 15 (BC) = 120


BC= 8

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