Theorem 62 sums up the pythagorean theorem: The square of the hypotenuse of a right triangle is equal to the sum of the squares of the legs.
We have been working with the pythagorean theorem for several years now, so there is no need to explain it further. We will be using this theorem in our problems today.
First, just a couple of new definitions:
1. pythagorean triplet - a group of 3 positive non-zero integers that satisify the pythagorean theorem, such as 3,4,5 and 12,35,37. If we plugged these numbers into a2 + b2 = c2, they will make a true statement.
2. constant - a quantity which remains the same throughout a given problem or discussion. The size of all angles inscribed in a given segment of a circle is constant, even though the vertex may change.
Corollary 62-1 - the difference of the square of the hypotenuse and the square of one leg equals the square of the other leg. This is just an algebraic manipulation of the formula.
Source: ABEKA Plane Geometry 2006
We will begin by proving theorem 62, the pythagorean theorem.
given, prove, and diagram in text, page 211
statements:
2. draw CD perpendicular to AB
3. AB:AC = AC:AD
AB:BC = BC:BD
4. (AD)2= AB X AD
(BC)2 = AB X BD
5. (AC)2 + (BC)2= AB X AD + AB X BD
6. (AC)2 + (BC)2= AB (AD + BD)
7. AB = AD + BD
8. (AC)2 + (BC)2 = (AB)2
Source: ABEKA Plane Geometry 2006
2. one and only one perpendicular can be drawn to a line from a point outside the line
3. if in a right triangle the perpendicular is drawn from the vertex of the right angle to the hypotenuse, each leg of the given triangle is the mean proportional between the hypotenuse and the adjacent segment
4. in any proportion the product of the means equals the product of the extremes
5. addition axiom
6. factoring
7. the whole is equal to the sum of its parts
8. substitution
Source: ABEKA Plane Geometry 2006
***Don't forget to use the diagram given in the text for theorem 62 with all of the exercises. Mark on it if needed, to help you see what sides are being discussed.
exercise 1:
If AC = 20 in and BC = 15 in, find AB
Source: ABEKA Plane Geometry 2006
You should have set your problem up like this:
(AB)2 = (AC)2 + (BC)2
Once solved, AB = 25 in.
Source: ABEKA Plane Geometry 2006
If AB = 60 in, AC = 45 in, find BC
Source: ABEKA Plane Geometry 2006
Be careful with this one. The original formula will need to be manipulated. It should look like this:
(BC)2 = (AB)2 - (AC)2
once solved, BC should equal 
= 15
in.
Source: ABEKA Plane Geometry 2006
If BC = 8 in, AB = 12 in, find AC.
in this problem, you are looking for the length of a side, not the hypotenuse
Source: ABEKA Plane Geometry 2006
Your formula should look like this:
(AC)2 = (AB)2 - (BC)2
AC should equal 
= 4
in.
Source: ABEKA Plane Geometry 2006
If BC = 24 in, BD = 14.4 in, find AB, AC, and DC
Take this one step by step. Look at the diagram, it is cut into two right triangles. Fill in the given info on the diagram and locate the hypotenuse and the required side. You may find it easier to set up proportions for some.
Source: ABEKA Plane Geometry 2006
First, solve for AB.
Your proportion should look like this:
BD:BC = BC:AB
*AB should equal 40 in
Second, solve for AC.
Your proportion should look like:
AD:AC = AC:AB
AB - BD:AC = AC:AB
AC should equal 32 in.
Third, solve for DC.
Your formula should look like this:
(DC)2 = (BC)2 - (BD)2
DC should equal 
= 19.2 in.
Source: ABEKA Plane Geometry 2006
If AC:BC = 3:4 and AB = 50 in, find AC and BC
Using the given proportion, let AC = 3x and BC = 4x
Your original formula would look like:
(AC)2 +(BC) = (AB)2
Now, substitute giving you:
(3x)2 + (4x)2 = 502
Source: ABEKA Plane Geometry 2006
AC = 3x = 30 in.
BC = 4x = 40 in.
Source: ABEKA Plane Geometry 2006
If AB:AC = 3:2 and BC = 35 in., find AB and AC
Following the principles of the previous problem, let AB = 3x and AC = 2x
Source: ABEKA Plane Geometry 2006
(AB)2 - (AC)2 = (BC)2
(3x)2 - (2x)2 = 352
AB = 3x = 21
AC = 2x = 14
Source: ABEKA Plane Geometry 2006
