Geometry Proofs: Chapter 7, Sections 7.1/7.2 Proof #1: We are given figures I and II as congruent right triangles with leg lengths a and b, and hypotenuse length c. So the area of figures I and II is ½ ab. We know that figure III is an isosceles right triangle since angles 1,2,3 make a straight angle while angles 1 and 3 are complementary (Corollary to Triangle Sum Th.) and thus angle 2 is a right angle. The area of figure III is ½ c 2 . We know that figures I,II,III are bordered by a trapezoid (Lines A to Transversal Th.) whose area is ½ (a + b)(a + b). By the Area Addition Postulate, the area of the trapezoid equals the sums of the areas of figures I,II,III: Proof #2: We are given figures I,II,III,IV as congruent right triangles with leg lengths a and b, and hypotenuse length c. So the area of figures I,II,III,IV is ½ ab. We know that figure V is a square (A Lines Form 4 Right Angles) with side lengths (b – a) with an area of (b – a) 2 . We know that figures I,II,III,IV,V are bordered by a square since angles 1 and 2 are complementary (Corollary to Triangle Sum Th.) and thus are right angles with side lengths of c whose area is c 2 . By the Area Addition Postulate, the area of the large square equals the sums of the areas of figures I,II,III,IV,V: Statements Reasons 1. ABC + with longest side c, c 2 = a 2 + b 2 1. 2. Construct Right PQR + with Right R and leg lengths a and b 2. Perpendicular & Segment Construction 3. 3. Pythagorean Theorem 4. 4. Substitution (#1,3) 5. 5. Property of Square Roots 6. ; ; AB PQ BC QR CA RP # # # 6. 7. ABC PQR # + + 7. 8. 8. 9. m C m R 9. 10. 90 m R q 10. 11. 11. 12. 12. Def. of Right Angle 13. 13. Def. of Right Triangle Pythagorean Theorem: Proof by Rearrangement of Area Given: Right triangle with leg lengths a and b, and hypotenuse length c. Prove: 2 2 2 c a b Converse of the Pythagorean Theorem: If the square of the length of the longest side of a triangle is equal to the sum of the squares of the lengths of the other two sides, then the triangle is a right triangle. Given: ABC + with longest side c, c 2 = a 2 + b 2 Prove: Right ABC + b a A B C c c b a I III II 1 2 3 b c a I V II III IV 2 1
Transcript
Geometry Proofs: Chapter 7, Sections 7.1/7.2
Proof #1: We are given figures I and II as congruent right triangles with leg lengths a and b, and hypotenuse length c. So the area of figures I and II is ½ ab. We know that figure III is an isosceles right triangle since angles 1,2,3 make a straight angle while angles 1 and 3 are complementary (Corollary to Triangle Sum Th.) and thus angle 2 is a right angle. The area of figure III is ½ c2. We know that figures I,II,III are bordered by a trapezoid (Lines A to Transversal Th.) whose area is ½ (a + b)(a + b). By the Area Addition Postulate, the area of the trapezoid equals the sums of the areas of figures I,II,III:
Proof #2: We are given figures I,II,III,IV as congruent right triangles with leg lengths a and b, and hypotenuse length c. So the area of figures I,II,III,IV is ½ ab. We know that figure V is a square (A Lines Form 4 Right Angles) with side lengths (b – a) with an area of (b – a)2. We know that figures I,II,III,IV,V are bordered by a square since angles 1 and 2 are complementary (Corollary to Triangle Sum Th.) and thus are right angles with side lengths of c whose area is c2. By the Area Addition Postulate, the area of the large square equals the sums of the areas of figures I,II,III,IV,V:
Statements Reasons 1. ABC+ with longest side c, c2 = a2 + b2 1.
2. Construct Right PQR+ with Right R� and leg lengths a and b
2. Perpendicular & Segment Construction
3. 3. Pythagorean Theorem
4. 4. Substitution (#1,3)
5. 5. Property of Square Roots
6. ; ;AB PQ BC QR CA RP# # # 6.
7. ABC PQR#+ + 7.
8. 8.
9. m C m R� � 9.
10. 90m R� q 10.
11. 11.
12. 12. Def. of Right Angle
13. 13. Def. of Right Triangle
Pythagorean Theorem: Proof by Rearrangement of Area Given: Right triangle with leg lengths a and b, and hypotenuse length c. Prove: 2 2 2c a b �
Converse of the Pythagorean Theorem: If the square of the length of the longest side of a triangle is equal to the sum of the squares of the lengths of the other two sides, then the triangle is a right triangle. Given: ABC+ with longest side c, c2 = a2 + b2 Prove: Right ABC+
b
a
A
B C
c c b
a I
III
II 1 2 3
b c
a I
V
II
III IV
2 1
Statements Reasons
1. ABC+ with longest side c, c2 < a2 + b2 1.
2. Construct right w/ Right and leg lengths and
PQRR a b
'�
2. Perpendicular & Segment Construction
3. ;BC QR CA RP# # 3.
4. 4. Pythagorean Theorem
2 25. c r� 5.
6. 6. Prop. of Square Roots
7. m C m R� � � 7.
8. 8. Def. of Right Angle
9. 9. Substitution (#7,8)
10. is acuteC� 10.
11. is an acute triangleABC' 11.
Statements Reasons
1. ABC+ with longest side c, c2 > a2 + b2 1.
2. Construct right w/ Right and leg lengths and
PQRR a b
'�
2. Perpendicular & Segment Construction
3. ;BC QR CA RP# # 3.
4. 4. Pythagorean Theorem
2 25. c r! 5.
6. 6. Prop. of Square Roots
7. m C m R� ! � 7.
8. 8. Def. of Right Angle
9. 9. Substitution (#7,8)
10. is obtuseC� 10.
11. is an obtuse triangleABC' 11.
Theorem 7.4: If the square of the length of the longest side of a triangle is less than the sum of the squares of the lengths of the other two sides, then the triangle is an acute triangle. Given: ABC+ with longest side c, c2 < a2 + b2 Prove: ABC+ is acute
Theorem 7.5: If the square of the length of the longest side of a triangle is greater than the sum of the squares of the lengths of the other two sides, then the triangle is an obtuse triangle. Given: ABC+ with longest side c, c2 > a2 + b2 Prove: ABC+ is obtuse
C A
B c
b
a
C A a c
b
B
Geometry Proofs: Chapter 7, Section 7.3
Statements Reasons 1. Right ABC+ with altitude CD 1.
2. 2. If the altitude is drawn to the hypotenuse of a right triangle, then the two triangles formed are similar to the original triangle and to each other.
3. AD CDCD BD
3.
Statements Reasons 1. Right ABC+ (with right ACB� ) with altitude CD
1.
2. 2. Def. of Altitude
3. ADC� , BDC� are right � ’s 3.
4. �ADC # �ACB, �BDC # �ACB
4.
5. 5. Reflexive Property of Congruence
6. +ACD~+ABC, +ABC~+CBD 6.
7. 7. Transitive Property of Similar Triangles
B
A
D
C
D
D
A C
C B
Th. #7.5: If the altitude is drawn to the hypotenuse of a right triangle, then the two triangles formed are similar to the original triangle and to each other. Given: Right ABC+ (with right ACB� ) with altitude CD Prove: ~ ~ABC ACD CBD+ + +
GMAT: The altitude drawn from the right angle to the hypotenuse of a right triangle is the geometric mean of the two hypotenuse parts. Given: Right ABC+ with altitude CD
Prove: AD CDCD BD
A
B C
D
Pythagorean Theorem (Using GMLT) In a right triangle, the square of the length of the hypotenuse is equal to the Sum of the squares of the lengths of the legs of the right triangle.
Given: Right ∆ABC, ∠BCA is a right angle
Prove: c2 = a2+b2
Statements Reasons
1. 1. Given
2. Draw CD ABA 2.
3. f a e b;a c b c 3.
4. cf=a2; ce=b2 4.
5. cf + ce = a2 + b2 5.
6. c = e + f 6.
7. c(f + e) = a2 + b2 7.
8. 8. Substitution (#6,7)
9. c2 = a2+b2 9.
Statements Reasons 1. Right ABC+ with altitude CD 1. 2. 2. If the altitude is drawn to
the hypotenuse of a right triangle, then the two triangles formed are similar to the original triangle and to each other.
3. AD ACAC AB
3.
GMLT: The leg of a right triangle is the geometric mean of the adjacent hypotenuse length when an altitude is drawn from the right angle to the hypotenuse and the whole hypotenuse. Given: Right ABC+ with altitude CD
