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# Proving Lines Proportional/Proportional Segments of Parallel Lines

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Author: Michele Harris
##### Description:

After studying sections 4.8 and 4.9, including theorem 60, the student will be complete the given proofs.

This lesson covers select exercises for  sections 4.8 and 4.9.

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Tutorial

## exercise #1, section 4.8

If ABC is a right triangle and CD is the altitude upon the hypotenuse, prove:

a) triangle ACD ~ triangle ABC

b) triangle CBD ~ triangle ABC

c) triangle ACD ~ triangle CBD

*draw and label your diagram

statements:

2.  CD perpendicular AB

3.  angle CDB, angle CDA are right angles

4.  triangle ACD and triangle CBD are right triangles

5. <B = <B,  <A = <A

6.  triangle ACD ~ triangle ABC

triangle CBD ~ triangle ABC

triangle ACD ~ triangle CBD

Source: ABEKA Plane Geometry 2006

## reasons for proof, exercise #1

​2.  definition of altitude

​3.  definition of perpendicular lines

​4.  definition of a right triangle

​5.  identity

​6.  two right triangles with a pair of acute angles = are ~.

Source: ABEKA Plane Geometry 2006

## exercise #2

​ABCD is a parallelogram and P any point on BC.  If AP and DC extended meet at O, prove:  triangle ABP ~triangle OCP ~triangle ODA

statements:

​2.  <O = <O

​3.  AD parallel to BC

AB parallel to OD

​4.  <D = <OCP

​5.  <B = <D

​6.  <D = <OCP = <B

​7. <PAB = <O

​8.  <O = <O = <PAB

​9.  triangle ABP ~ triangle OPC ~ triangle OAD

Source: ABEKA Plane Geometry 2006

## Reasons for proof, exercise #2

​2.  identity

​3.  definition of parallelogram

​4.  if 2 parallel lines are cut by transversal, then the corresponding angles formed are equal.

​5.  opposite angles of a parallelogram are equal.

​6.  substitution

​7.  if 2 parallel lines are cut by a transversal, then the alternate interior angles formed are equal.

​8.  substitution

​9.  a.a.

Source: ABEKA Plane Geometry 2006

## exercise #5

​The diagonals of a trapezoid divide each other proportionally.

​Given:  trapezoid ABCD with diagonal AC, BD intersecting at O

​Prove:  AO:CO = BO:DO

​statements:

​2.  AB parallel to CD

​3.  <CDB = <ABD

<DCA = <BAC

​4.  triangle AOB ~ triangle COD

​5.  AO:CO = BO:DO

Source: ABEKA Plane Geometry 2006

## Reasons for proof, exercise #5

​2.  definition of a trapezoid

​3.  if two parallel lines are but by a transversal, then the alternate interior angles are equal.

4.  a.a.

​5.  C.S.S.T.P.

Source: ABEKA Plane Geometry 2006

## exercise #1, section 4.9

​From the trisection points of side AB of triangle ABC lines are drawn to C.  Prove that these lines trisect all lines parallel to AB and are terminated by the other two sides.

​given: triangle ABC

D, E trisection points of AB

CD, CE

RU parallel to AB

prove:  CD and CE trisect RU

statements:

2.  AD:RS = DE:ST = EB:TU

3.  AD = DE = EB

6.  RS = ST - TU

7.  CD and CE trisect RU

Source: ABEKA Plane Geometry 2006

## Proof for exercise #1, section 4.9

​2.  if 2 parallels are cut by three or more transversals passing through a common point, the corresponding segments of the parallels are proportional

​3.  definition of the trisection of a segment

​4.  substitution axiom

​5.  inversion transformation

​6.  multiplication axiom (multiply each term by AD which then cancels)

​7. definition of trisector

Source: ABEKA Plane Geometry 2006