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
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
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
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
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
The fundamental property of proportions: the product of the means equals the product of the extremes
Source: ABEKA Plane Geometry 2006
***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
AD : DC = DC : DB
9 : DC = DC : 4
(DC)2 = 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
If DC = 8 in, AD = 16 in, find DB, BC, AND AC
Source: ABEKA Plane Geometry 2006
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
If AC = 6 in, AD = 4 in, find AB and BC
Source: ABEKA Plane Geometry 2006
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
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
6/15 = 2/5; 8/20 = 2/5; 10/25 = 2/5
yes, since the corresponding sides are proportional
Source: ABEKA Plane Geometry 2006
