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1. Construct AB.
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2. Construct external point P not on AB.
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3. Draw a transversal through P that intersects AB at R.
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4. Using compass, make R the center of a circle of radius less than . Draw an arc.
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5. Keep the compass with the same opening and draw an intersecting arc from point Q.
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6. Repeat at point P and then S.
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7. SPD Å QRB.
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8. Therefore CD||AB.
(figure available in print form)
Equilateral Triangle
1. Make line AB of any convenient length.
2. Open a compass to the length of AB. _
3. Use A as the center of a circle of radius AB.
4. Use B as the center of a second circle also of radius AB.
5. Where the two circles intersect at the top, label point C.
6. Draw lines AC and BC.
7. AC, AB, and BC are all radii of equal circles.
8. AC Å AB Å BC, therefore ’ ABC is equilateral.
(figure available in print form)
Construction of Isosceles Triangle
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1. Draw line BC.
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2. Open compass to length BC.
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3. Use point B as the center of a circle of radius BC.
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4. Repeat #3, but use C as the center.
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5. Extend line BC to new length AD.
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6. Open compass to this new length.
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7. Make two new circles of radii AD using B and C as the center points.
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8. Where the two larger circles intercept label it point E
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9. Draw lines BE, CE, and BC.
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10. BC is the radius of one smaller circle.
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11. BE and CE are radii of congruent circles.
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12. Therefore BE Å CE.
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13. BE > BC. 14. ’ BCE is Isosceles per definition.
(figure available in print form)
SCALENE TRIANGLE
1. Draw line AB of any length.
2. Open compass to equal the length of AB.
3. Using A, and then B, as the centers, draw two circles of radius AB.
4. Put a point on the circumference of one circle and label it point C.
5. Draw radius AC.
6. Choose a point D on radius AC such that AD AC.
7. Draw a line from point B to point D making line BD.
8. AD AB.
9. BD > AB.
10. BD > AD.
11. Therefore, all lines are of different length, resulting in a Scalene Triangle.
(figure available in print form)
Proof of two parallel lines
Given: AB is a line, 5 Å $l t7 Prove: CD AB
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Statement
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Reason
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EF is a transversal.
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1. Definition of Transversal.
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2. 5 = 7
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2. Given
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3. 6 Å 8
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3. Opposite angles
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4. 5 Å 3
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4. Alternate Interior angles
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5. 3 Å 1
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5. Opposite angles
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6. 5 Å 1
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6. Substitution
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7. CD |AB
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7. Two lines are parallel when a Transversal forms congruent alternate interior angles
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(figure available in print form)
Proof of Perpendicular Lines
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Given: AB bisects CD
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Prove: AB CD
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Statement
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Reason
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1. AB bisects CD
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1. Given
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2. CB Å BD
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2. Definition of bisector
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3. Construct lines AC and AD
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3. Construction
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4. ACÅAD
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4. Definition of bisector
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5. AB = AB
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5. Reflexive property
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6. ’ ABC Å’ BD
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6. Side, side, side theorem
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7. Choose a point E on AB
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7. Construction
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and construct CE and DE
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8. CE Å DE
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8. Definition of bisector
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9. EBÅ EB
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9. Reflexive property
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10. ’ BCEÅ’BDE
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10. Same as 6
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11. AB CD
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11. A line determined by two points equidistant from the ends of a line segment is the perpendicular bisector of the line segment.
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(figure available in print form)
Proof of an Equilateral Triangle:
Given: AB Å BC, B Å C Prove: ’ABC is equilateral.
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Statement
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Reason
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1. AB Å BC
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1. Given
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2. B Å C
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2. Given
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3. A Å C
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3. Angles opposite equal sides are equal.
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4. A Å B
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4. Substitution
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5. AC Å AB
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5. Sides opposite equal angles are equal.
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6. AC Å BC
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6. Substitution
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7. ’ ABC is equilateral
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7. Definition of equilateral
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(figure available in print form)
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Given: ’ ABC with A Å B, CA Å CB
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Prove: ’ ABC is Isosocles
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Statement
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Reason
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1. A Å B
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1. Given
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2. Draw BD such that B is
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2. Construction and definition of
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bisected
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bisected
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3. 1 Å 2
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3. Definition of bisector
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4. BD ÅBD
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4. Reflexive property
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5. ’ ABDÅ ’CBD
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5. Side, angle, angle theorem
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6. ABÅ BC
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6. Corresponding sides of congruent angles are congruent.
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7. ’ ABC is Isosocles
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7. Definition of Isosocles
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(figure available in print form)
Proof of a Scalene Triangle
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Given: A > C, BC AC
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Prove: ’ ABC is Scalene
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Statement
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Reason
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1. A > C
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1. Given
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2. BC AC
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2. Given
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3. C A
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3. Based on 1
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4. AB BC
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4. Side opposite C is less than the side opposite A (based on corresponding sides opposite their angles)
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5. AC > BC
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5. Based on 2
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6. ’ABC is Scalene
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6. Definition of Scalene Triangle
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(figure available in print form)
THIS PHOTO OF THE BACK COURTYARD OF WOOLSEY HALL HAS SEVERAL ELEMENTS. FIRST, SOME INTANGIBLE ASPECTS. THE SPACE SEEMS ARRANGED IN A PLEASANT MANNER. THERE ARE NO SHARP CONTRASTS TO SHOCK US. THERE ARE THE PLEASANT CURVING LINES OF THE WINDOWS WHILE THE COLUMNS PRESENT THE USE OF PERSPECTIVE ALLOW FOR GRADUAL CHANGE. THESE CHARACTERISTICS ARE TOWARD THE BEAUTIFUL AND GOING TO A CONCERT HALL WOULD PROBABLY BE A BEAUTIFUL EXPERIENCE. CONSIDERING THE COLUMNS ON THE LEFT AND ALSO ANOTHER VIEW OF THE SAISE COLUMNS, WHAT TANGIBLE QUALITIES CAN YOU FIND? WHAT SORT OF GEOMETRIC SHAPE ARE THE COLUMNS? WHAT KIND OF GEOMETRIC CHARACTERISTIC IS PRESENT IN ONE COLUMN NEXT TO ANOTHER? CAN YOU DISCERN ANYTHING ELSE?
(figure available in print form)
(figure available in print form)
IF YOU WERE ENTERING A BUILDING SURROUNDED BY A FACE LIKE THIS, WHAT EFFECT WOULD THAT HAVE ON YOU? ENTWINING SNAKES: THE UNINTERRUPTED REPETITION OF SPIKED POLES: THE SHARP CONTRAST OF A BLACK FENCE AGAINST A WHITE STONE BACKGROUND? UNCOMFORTABLE?
(figure available in print form)
ANOTHER EXAMPLE OF THE INTANGIBLE. THIS SPACE HAS A COMFORTING EFFECT. PLEASANT GREEN COLORS, BROWNS, AND OTHER EARTH TONES. PLENTY OF SPACE AROUND EACH BENCH AND THE GENTLE ARCH OF THE TREES.
(figure available in print form)
AN EXAMPLE OF THE INTANGIBLE. A QUIET CORNER. A GRADUAL CHANGE AS YOU COME THROUGH A COURTYARD INTO A WELL LIGHTED AREA. A GENTLENESS IS EXPERIENCED BECAUSE OF SHRUBS TAKING THE SHARP EDGE OFF THE BUILDING.
(figure available in print form)
EXAMPLE OF A CUBE—THE BEINECKE RACE BOOKS LIBRARY. THE CUBE ITSELF IS A BASIC GEOMETRIC SHAPE. CAN YOU FIND LARGE SQUARES OR RECTANGLES? WHAT OTHER SHAPES CAN YOU FIND
(figure available in print form)
HERE IS AN EXAMPLE OF USE OF THE GOLDEN SECTION OF GREEK ARCHITECTURE. THIS IS THE OLD POST OFFICE DOWNTOWN. ALLOWING FOR AN INACCURACY DUE TO PHOTO DISTORTION, WHEN YOU MEASURE FROM THE OUTER EDGE OF THE EXTREME LEFT HAND COLUMN ALL THE WAY TO THE OUTER EDGt OF THE EXTREME RIGHT HAND COLUMN, YOU GET 3 1/16 INCHES OR 3.0625 INCHES. THEN MEASURE FROM THE VERTEX OF THE FRIEZE TO THE BASE OF THE COLUMNS AND YOU GET 1 29/32 INCHES OR 1.9063. DIVIDE 3.06Z5 BY 1.9063 AND YOU CONE UP WITH 1.6065. THE GOLDEN SECTION IS 1.61. THIS WAS CONSIDERED THE MOST PLEASING PROPORTION IN GREEK CULTURE AND AN EXPRESSION OF BEAUTY. AS DISCUSSED EARLIER, EDMUND BURKE RELATES THE HIGH IDEALS OF WISDOM, TRUTH, JUSTICE, ETC., TO THE SENSE OF BEAUTY. GOVERNMENTS REPRESENTED SUCH HIGH IDEALS. THUS WE HAVE A POST OFFICE DESIGNED USING THE GOLDEN SECTION.
(figure available in print form)
THIS IS THE COURT HOUSE ON THE NEW HAVEN GREEN. THE—EDGE SECTION IS PRESENT IN THIS GOVERNMENT BUILDING. MEASURE THE WIDTH OF THE COLUMNS AT THEIR BASE AND YOU GET 4 7/8 IN. OR 4.875 IN. THEN MEASURE THE LENGTH OF THE COLUMNS THEMSELVES AND YOU GET AN EVEN 3 INCHES 4.875: 3.0 = 1.625. ALLOWING FOR PHOTO DISTORTION, YOU HAVE THE GOLDEN RATIO OF 1:1.61.
(figure available in print form)
