+
Proportional Segments in Right Triangles

Proportional Segments in Right Triangles

Author: Michele Harris
Description:

The student wil be able to use given lengths to tell if sides of right triangles are proportional,

Exercises for section 4.10

(more)
See More

Try Our College Algebra Course. For FREE.

Sophia’s self-paced online courses are a great way to save time and money as you earn credits eligible for transfer to over 2,000 colleges and universities.*

Begin Free Trial
No credit card required

25 Sophia partners guarantee credit transfer.

221 Institutions have accepted or given pre-approval for credit transfer.

* The American Council on Education's College Credit Recommendation Service (ACE Credit®) has evaluated and recommended college credit for 20 of Sophia’s online courses. More than 2,000 colleges and universities consider ACE CREDIT recommendations in determining the applicability to their course and degree programs.

Tutorial

Theorem 61

​If in a rightr triangle the perpendicular is drawn from the vertex of the right angle to the hypotenuse:

​1.  The two triangles thus formed are similar to the given triangle and to each other

​2.  The perpendicular is the mean proportional between the segments of the hypotenuse

​3.  Each leg of the given triangle is the mean proportional between the hypotenuse and the adjacent segment

 

​given: right triangle ABC, and CD the perpendicular from C to the hypotenuse AB

​prove:  triangle ADC ~ triangle ABC ~ triangle DBC

           AD : DC = DC : DB

          AD : AC = AC : AB AND DB : BC = BC : AB       

 

      statements for proof:

​2.  <C = right <

​3.  AC perpendicular BC

​4.  <A = < x prime, <B = < y prime

​5.  <A = <A, <B = <B

​6.  triangle ADC ~ triangle ABC ~ triangle DBC

​7.  AD : DC = DC : BD

    AD : AC = AC : AB

   DB : BC = BC : AB

Source: ABEKA Plane Geometry 2006

Reasons for proof, theorem 61

2.  definition of right triangle

​3.  definition of perpendicular lines

​4.  Angles  whose sides are perpendicular right to right and left to left are =

​5.  identity

 

6.  a.a.

​7.  C.S.S.T.P.

Source: ABEKA Plane Geometry 2006

Exercise #1

In circle O, PA is a tangent and AB a diameter.  If PB cuts the circle at C, prove (AC)2 = BC x CP.

*Use  diagram given in text.

​given:  circle O, diameter AB; tangent PA; secant PB cutting circle O in C

​prove:  (AC)2 = BC x CP

 

​statements for proof:

​2.  draw AC

​3.  arc ACB is a semicircle

​4.  <ACB is a right angle

​5.  AC perpendicular to PB

​6.  PA perpendicular to AB

​7.  <PAB is a right angle

​8.  triangle PAB is a right triangle

​9.  BC : AC = AC : CP

​10.  (AC)​2​ = BC X CP

 

Source: ABEKA Plane Geometry 2006

Reasons for proof, #1

​2.  through 2 given points one and only one straight line can be drawn

​3.  a diameter bisects a circle into two semicircles

​4.  an angle inscribed in a semicircle is a right angle

5.  definition of perpendicular lines

​6.  a tangent to a circle is perpendicular to the radius at the point of contact

​7.  definition of perpendicular lines

​8.  definition of a right triangle

​9.  if in a right triangle the perpendicular is drawn from the vertex of the right angle to the hypotenuse, the perpendicular is the mean proportional between the segments of the hypotenuse

​10.  in any proportion the product of the means equals the product of the extremes

Source: ABEKA Plane Geometry 2006

Exercise #3

How can it be shown in Theorem 61 that (DC)2 = AD x DB?  (AC)​2 = AD x AB?

(BC)​2​ = DB x AB?

* use the diagram given in the text

Source: ABEKA Plane Geometry 2006

answer for #3

​The fundamental property of proportions:  the product of the means equals the product of the extremes

Source: ABEKA Plane Geometry 2006

Exercise #4

***Use the drawing for Theorem 61 for exercises 4 through 6.  Express answers as radicals and fractions if applicable.

 

​If AD = 9 in, DB = 4 in, find DC, AC, and BC

Source: ABEKA Plane Geometry 2006

answer for #4

AD : DC = DC : DB

​9 : DC = DC : 4

(DC)= 36

​DC = 6 in.

 

AD : AC = AC : AB

AD: AC = AC : AD + DB

​9 : AC = AC : 13

(AC)2= 117

​AC = square root of 117 = 10.8 in.

Source: ABEKA Plane Geometry 2006

exercise #5

​If DC = 8 in, AD = 16 in, find DB, BC, AND AC

Source: ABEKA Plane Geometry 2006

answer for #5

​AD : DC = DC : DB

​16 : 8 = 8 : DB

​16DB = 64

​DB = 4 in. 

 

​DB : BC = BC : AB

​DB : BC = BC : AD + DB

​4 : BC = BC : 20

(BC)2= 80

​BC = square root of 80 = 8.9 in.

 

​AD : AC = AC : AB

​16 : AC = AC : AD + DB

​16 : AC = AC : 20

(AC)2 = 320

​AC = square root of 320 = 17.9 in.

 

Source: ABEKA Plane Geometry 2006

exercise #6

If AC = 6 in, AD = 4 in, find AB and BC

Source: ABEKA Plane Geometry 2006

answer for #6

​AD : AC = AC : AB

​4 : 6 = 6 : AB

​4AB = 36

​AB = 9 in.

 

​DB : BC = BC : AB

​AB - AD:BC = BC:AB

​5 : BC = BC : 9

(BC)​2​ = 45

​BC = square root of 45 = 6.7 in.

Source: ABEKA Plane Geometry 2006

exercise #11

If the sides of two triangles are 6 in, 8 in, 10 in, and 15 in, 20 in, 25 in. respectively, are the two triangles similar?

Source: ABEKA Plane Geometry 2006

answer for # 11

​6/15 = 2/5; 8/20 = 2/5;  10/25 = 2/5

 

​yes, since the corresponding sides are proportional

Source: ABEKA Plane Geometry 2006