Practice with Two-Column Proofs

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PRACTICE WITH TWO-COLUMN PROOFS

Jump right to the exercises!

In this section, you will practice with two-column proofs involving the Pythagorean Theorem, triangle congruence theorems, and other tools.

We start with a definition:

DEFINITION (geometric mean)

Given two positive numbers a and b ,
the geometric mean of a and b is the positive number ab .

Equivalently, the geometric mean of a and b
is the positive solution g of the equation ag =gb .
(Solve for g and take the positive solution.)

This number g is also referred to as the geometric mean between a and b .

A moment's reflection shows that if a and b are the lengths of the sides of a rectangle,
then the geometric mean of a and b gives the side length of the square with the same area as the rectangle.

geometric mean between a and b

The next proof gives an interesting result:
if the altitude of a triangle is the geometric mean between the two segments of the side it hits,
then the triangle must be a right triangle.

The proof can be written in different orders.
In the exercises, you will practice supplying reasons for each step in the proof.

GIVEN: CB¯ ⊥AD ¯ ABBC =BCBD PROVE: ∠ACD is a right angle
STRATEGY: Show that (AC) 2+(CD )2 =(AB+BD) 2
PROOF:
STATEMENTS	REASONS
1. CB¯ ⊥AD ¯	given
2. ΔABC is a right triangle with hypotenuse AC¯	definitions: right triangle, hypotenuse
3. (AB) 2+(BC )2 =(AC) 2	the Pythagorean Theorem
4. ΔDBC is a right triangle with hypotenuse CD¯	definitions: right triangle, hypotenuse
5. (BD) 2+(BC )2 =(CD) 2	the Pythagorean Theorem
6. (AC) 2+(CD )2 =(AB) 2+(BC )2+ (BD) 2+(BC )2	substitution (steps 3 and 5)
7. (AC) 2+(CD )2 =(AB) 2+2(BC )2+ (BD) 2	combine like terms
8. ABBC =BCBD	given
9. (AB) (BD)= (BC) 2	Multiplication Property of Equality; multiply both sides by (BC)(BD ) (cross-multiply)
10. 2(AB) (BD)=2 (BC) 2	Multiplication Property of Equality; multiply both sides by 2
11. (AC) 2 + (CD) 2 = (AB) 2 + 2 (AB) (BD) + (BD) 2	substitution (steps 7 and 10)
12. (AB+B D)2= (AB)2 +2(AB) (BD)+ (BD) 2	FOIL
13. (AC) 2+(CD )2 =(AB+BD) 2	substitution (steps 11 and 12)
14. AB+BD= AD	B is between A and D
15. (AC) 2+(CD )2= (AD) 2	substitution (steps 13 and 14)
16. ΔACD is a right triangle with hypotenuse AD¯ ; so, ∠ACD is a right angle	the Pythagorean Theorem (converse)

This next proof can be done in different ways.
In the exercises, you will practice supplying reasons for each step in the proof.

GIVEN: AB= AC D is the midpoint of AB¯ E is the midpoint of AC¯ PROVE: DF= EF
STRATEGY: Show that ΔBDF≅ ΔCEF
PROOF:
STATEMENTS	REASONS
1. AB= AC D is the midpoint of AB¯ E is the midpoint of AC¯	given
2. m∠ACB= m∠ABC	angles opposite equal sides have equal measures
3. BD=1 2AB and CE=1 2AC	definition of midpoint
4. BD=CE	substitution (steps 1 and 3)
5. BC¯≅CB ¯	reflexive property
6. ΔDBC≅ ΔECB	SAS (steps 2, 4, 5)
7. m∠BCD =m∠ CBE	CPCTC
8. BF= CF	sides opposite equal measure angles are equal
9. m∠ABE +m∠CBE =m∠ ABC and m∠ACD +m∠BCD =m∠ ACB	angle addition
10. m∠ABE +m∠BCD =m∠ ACD +m∠BCD	substitution (steps 2, 7, and 9)
11. m∠ABE =m∠ACD	addition property of equality (subtract m∠BCD from both sides)
12. ∠DFB≅ ∠EFC	vertical angles are congruent
13. ΔBDF≅ ΔCEF	ASA (steps 8, 11, 12)
14. DF=EF (or, DF¯ ≅EF¯ )	CPCTC

In the exercises, you will also practice doing shorter two-column proofs;
you will need to supply both the statements and reasons.

In your proofs, feel free to use statements like AB¯ ≅CD ¯ and AB=CD totally interchangeably,
since two segments are congruent if and only if they have the same length.
However, be careful to use the verb ≅ to compare geometric figures, and the verb = to compare numbers.

Similarly, feel free to use statements like ∠ABC≅∠ DEF and m∠ABC =m∠DEF totally interchangeably,
since two angles are congruent if and only if they have the same measure.

On this exercise, you will not key in your answer.
However, you can check to see if your answer is correct.