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# Proportional Segments in Right Triangles

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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

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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 = = 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 = = 8.9 in.

​AD : AC = AC : AB

​16 : AC = AC : AD + DB

​16 : AC = AC : 20

(AC)2 = 320

​AC = = 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 = = 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

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