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Parallelogram Exactly 1 pair of parallel sides 2 pairs of parallel sides Exactly 1 pair of congruent sides 2 pairs of congruent sides All sides congruent Perpendicular diagonals Congruent diagonals Diagonals bisect angles Diagonals bisect each other Opposite angles congruent Four right angles Base angles congruent Rhombus Rectangle Square Trapezoid Isosceles Trapezoid Parallelogram Rhombus Rectangle Square Trapezoid Isosceles Trapezoid Exactly 1 pair of What it take to prove that a quadrilateral is a rhombus? parallel does sides 2 pairsWe of parallel know a rhombus has perpendicular diagonals. sides Does this mean that if we have a quadrilateral with perpendicular Exactly 1 pair of congruent sides diagonals, it must be a rhombus? 2 pairs of congruent sides All sides congruent Perpendicular diagonals Congruent diagonals Diagonals bisect angles Diagonals bisect each other Opposite angles congruent Four right angles Base angles congruent Ways to prove that a quadrilateral is a parallelogram Ways to prove that a quadrilateral is a parallelogram Show both pairs of opposite sides parallel Ways to prove that a quadrilateral is a parallelogram Show both pairs of opposite sides parallel Show one pair of opposite sides parallel and congruent C D A B Ways to prove that a quadrilateral is a parallelogram Show both pairs of opposite sides parallel Show one pair of opposite sides parallel and congruent Show both pairs of opposite sides congruent Show both pairs of opposite angles congruent Show that the diagonals bisect each other Ways to prove that a quadrilateral is a parallelogram Show both pairs of opposite sides parallel Show one pair of opposite sides parallel and congruent Show both pairs of opposite sides congruent Show both pairs of opposite angles congruent Show that the diagonals bisect each other Ways to prove that a quadrilateral is a rhombus parallelogram and two adjacent sides congruent Ways to prove that a quadrilateral is a parallelogram Show both pairs of opposite sides parallel Show one pair of opposite sides parallel and congruent Show both pairs of opposite sides congruent Show both pairs of opposite angles congruent Show that the diagonals bisect each other Ways to prove that a quadrilateral is a rhombus parallelogram and two adjacent sides congruent parallelogram and perpendicular diagonals C D A B Ways to prove that a quadrilateral is a parallelogram Show both pairs of opposite sides parallel Show one pair of opposite sides parallel and congruent Show both pairs of opposite sides congruent Show both pairs of opposite angles congruent Show that the diagonals bisect each other Ways to prove that a quadrilateral is a rhombus parallelogram and two adjacent sides congruent parallelogram and perpendicular diagonals parallelogram and diagonals bisect its angles Ways to prove that a quadrilateral is a parallelogram Show both pairs of opposite sides parallel Show one pair of opposite sides parallel and congruent Show both pairs of opposite sides congruent Show both pairs of opposite angles congruent Show that the diagonals bisect each other Ways to prove that a quadrilateral is a rhombus parallelogram and two adjacent sides congruent parallelogram and perpendicular diagonals parallelogram and diagonals bisect its angles Ways to prove that a quadrilateral is a rectangle parallelogram and one right angle parallelogram and congruent diagonals Ways to prove that a quadrilateral is an isosceles trapezoid trapezoid and non-parallel sides congruent trapezoid and one pair of congruent base angles trapezoid and congruent diagonals B We know that an altitude of a triangle is a segment from a vertex that is perpendicular to the opposite side (AD in the diagram). D A The median of a triangle is a segment from a vertex to the midpoint of the opposite side. A B C M AM is a median of triangle ABC. C A MN is a midsegment of triangle ABC N M C B A MN is the median or midsegment of trapezoid ABCD M D B N C The segment joining the midpoints of two sides of a triangle (the midsegment) is parallel to the third side and half its length. A ½x M N x C B The median of a trapezoid is parallel to the bases and is equal in length to the average of the lengths of the bases. A M D b1 ½ (b1 + b2) B N b2 C The segment joining the midpoints of two sides of a triangle (the midsegment) is parallel to the third side and half its length. A M B N C P The segment joining the midpoints of two sides of a triangle (the midsegment) is parallel to the third side and half its length. AM PC, A PCA (CPCTC) ( alternate interior angles) MP BC A M B N C P The segment joining the midpoints of two sides of a triangle (the midsegment) is parallel to the third side and half its length. AM PC, A PCA (CPCTC) ( alternate interior angles) MP BC A M B N C P The segment joining the midpoints of two sides of a triangle (the midsegment) is parallel to the third side and half its length. AM PC, A PCA (CPCTC) ( alternate interior angles) MP BC A M B N C P The segment joining the midpoints of two sides of a triangle (the midsegment) is parallel to the third side and half its length. AM PC, A PCA (CPCTC) ( alternate interior angles) MP BC A M B N C P The segment joining the midpoints of two sides of a triangle (the midsegment) is parallel to the third side and half its length. AM PC, A PCA (CPCTC) ( alternate interior angles) MP BC A M B N C P The segment joining the midpoints of two sides of a triangle (the midsegment) is parallel to the third side and half its length. AM PC, A PCA (CPCTC) ( alternate interior angles) MP BC A M B N C P The segment joining the midpoints of two sides of a triangle (the midsegment) is parallel to the third side and half its length. AM PC, A PCA (CPCTC) ( alternate interior angles) MP BC A M B N C P The median of a trapezoid is parallel to the bases and is equal in length to the average of the lengths of the bases. A M D B N C The median of a trapezoid is parallel to the bases and is equal in length to the average of the lengths of the bases. Proof: A M D B P N C Because MP is the midsegment of ABD, MP is parallel to AB by the previous theorem. The median of a trapezoid is parallel to the bases and is equal in length to the average of the lengths of the bases. Proof: A M D B P N C Because MP is the midsegment of ABD, MP is parallel to AB by the previous theorem. Since MP is part of MN, MN is parallel to AB. The median of a trapezoid is parallel to the bases and is equal in length to the average of the lengths of the bases. Proof: A M D B P N C Because MP is the midsegment of ABD, MP is parallel to AB by the previous theorem. Since MP is part of MN, MN is parallel to AB. Similarly, PN is the midsegment of ADC, so that PN is parallel to DC. Since PN is part of MN, MN is parallel to DC. Therefore, median MN is parallel to bases AB and CD. The median of a trapezoid is parallel to the bases and is equal in length to the average of the lengths of the bases. Proof: A M D B P N C Because MP is the midsegment of ABD, MP is parallel to AB by the previous theorem. Since MP is part of MN, MN is parallel to AB. Where is the flaw in this proof? Similarly, PN is the midsegment of ADC, so that PN is parallel to DC. Since PN is part of MN, MN is parallel to DC. Therefore, median MN is parallel to bases AB and CD. The median of a trapezoid is parallel to the bases and is equal in length to the average of the lengths of the bases. Proof: A M D B N C First we prove the following lemma: If two lines are parallel to the same line, they are parallel to each other. If m // l and n // l, prove m // n l The median of a trapezoid is parallel to the bases and is equal in length to the average of the lengths of the bases. Proof: A M D B N C First we prove the following lemma: If two lines are parallel to the same line, they are parallel to each other. If m // l and n // l, prove m // n l The median of a trapezoid is parallel to the bases and is equal in length to the average of the lengths of the bases. Proof: A M D B N C First we prove the following lemma: If two lines are parallel to the same line, they are parallel to each other. If m // l and n // l, prove m // n l The median of a trapezoid is parallel to the bases and is equal in length to the average of the lengths of the bases. Proof: A M B P D MP is parallel to AB. It is a midsegment of ABD PN is parallel to DC. It is a midsegment of BDC N C DC is parallel to AB. Definition of trapezoid Therefore, by the lemma, MP is parallel to DC. (they are both parallel to AB) But Euclid’s Parallel Postulate says there can only be one line through P parallel to DC. Therefore,MN PN and MP must be same line, i.e. DC. MN. Therefore, is parallel tothe both AB and The median of a trapezoid is parallel to the bases and is equal in length to the average of the lengths of the bases. Proof: A M D b1 ½ b1P B ½ b2 N b2 MN = ½ b1 + ½ b2 = ½ (b1 + b2) C The segment joining the midpoints of two sides of a triangle (the midsegment) is parallel to the third side and half its length. A ½x M N x C B The median of a trapezoid is parallel to the bases and is equal in length to the average of the lengths of the bases. A M D b1 ½29(binches 1 + b2 ) B N 42binches 2 C The altitude of a parallelogram C B The altitude of a trapezoid D A D A C B The following construction problems should be worked on and solved in groups of three people. Each group will be assessed three times— once for each category. When your group is ready to be assessed, notify me and indicate on which category the group wants to be assessed. Your group will choose the problem in that category on which to be assessed. The group member who will be asked to solve the problem will be selected randomly. Whatever grade is assigned to that student will be the grade for the group. Be sure each group member can solve each problem correctly. The following scoring guidelines will be used to assess each problem. The maximum possible point total is 45 points. Construction 0 Cannot do the construction correctly even with hints 3 Can do the construction correctly but only after hints are given 6 Can do the construction correctly without hints The following construction problems should be worked on and solved in groups of three people. Each group will be assessed three times— once for each category. When your group is ready to be assessed, notify me and indicate on which category the group wants to be assessed. Your group will choose the problem in that category on which to be assessed. The group member who will be asked to solve the problem will be selected randomly. Whatever grade is assigned to that student will be the grade for the group. Be sure each group member can solve each problem correctly. The following scoring guidelines will be used to assess each problem. The maximum possible point total is 45 points. Proof 0 Cannot do the proof correctly even with hints 3 Can do the proof correctly but only after hints are given 6 Can do the proof correctly without hints Sample: Construct a rhombus given one side and an altitude 1. Construct a rhombus (not a square), given a. one side and one angle (1 point) b. one angle and a diagonal (2 points) c. the altitude and one diagonal (3 points) 2. Construct a parallelogram (not a rhombus), given a. one side, one angle, and one diagonal (1 point) b. two adjacent sides and an altitude (2 points) c. one angle, one side, and the altitude on that side (3 points) 3. Construct an isosceles trapezoid, given a. the diagonal, altitude, and one of the bases (1 point) b. one base, the diagonal, and the angle included by them (2 points) c. the bases and one angle (3 points) Construct a rhombus given one side and an altitude. Construct a rhombus given one side and an altitude. Solution 1. Begin by copying one of the sides on the “Figures” sheet and label it AB. Construct a rhombus given one side and an altitude. Solution 1. Begin by copying one of the sides on the “Figures” sheet and label it AB. 2. Choose a point P between A and B Construct a rhombus given one side and an altitude. Solution 1. Begin by copying one of the sides on the “Figures” sheet and label it AB. 2. Choose a point P between A and B and construct a perpendicular to AB through P. Construct a rhombus given one side and an altitude. Solution 1. Begin by copying one of the sides on the “Figures” sheet and label it AB. 2. Choose a point P between A and B and construct a perpendicular to AB through P. 3. Copy the altitude from the “figures” sheet so that one its endpoints is point P and the other is on the perpendicular you constructed. Label the other endpoint Q. Construct a rhombus given one side and an altitude. Solution 1. Begin by copying one of the sides on the “Figures” sheet and label it AB. 2. Choose a point P between A and B and construct a perpendicular to AB through P. 3. Copy the altitude from the “figures” sheet so that one its endpoints is point P and the other is on the perpendicular you constructed. Label the other endpoint Q. 4. Using point A as center, make a circle with radius AB. Construct a rhombus given one side and an altitude. Solution 1. Begin by copying one of the sides on the “Figures” sheet and label it AB. 2. Choose a point P between A and B and construct a perpendicular to AB through P. 3. Copy the altitude from the “figures” sheet so that one its endpoints is point P and the other is on the perpendicular you constructed. Label the other endpoint Q. 4. Using point A as center, make a circle with radius AB. 5. Construct a line through point Q perpendicular to PQ (construction marks not shown). Label the point where it intersects the circle point R. Construct a rhombus given one side and an altitude. Solution 1. Begin by copying one of the sides on the “Figures” sheet and label it AB. 2. Choose a point P between A and B and construct a perpendicular to AB through P. 3. Copy the altitude from the “figures” sheet so that one its endpoints is point P and the other is on the perpendicular you constructed. Label the other endpoint Q. 4. Using point A as center, make a circle with radius AB. 5. Construct a line through point Q perpendicular to PQ (construction marks not shown). Label the point where it intersects the circle point R. 6. Construct radius AR. Construct a rhombus given one side and an altitude. Solution 1. Begin by copying one of the sides on the “Figures” sheet and label it AB. 2. Choose a point P between A and B and construct a perpendicular to AB through P. 3. Copy the altitude from the “figures” sheet so that one its endpoints is point P and the other is on the perpendicular you constructed. Label the other endpoint Q. 4. Using point A as center, make a circle with radius AB. 5. Construct a line through point Q perpendicular to PQ (construction marks not shown). Label the point where it intersects the circle point R. 6. Construct radius AR. 7. Using point R as center, mark off a length equal to AB along ray QR. Call the intersection point T. Construct a rhombus given one side and an altitude. Conclusion: ARTB is a rhombus with the required information. Solution 1. Begin by copying one of the sides on the “Figures” sheet and label it AB. 2. Choose a point P between A and B and construct a perpendicular to AB through P. 3. Copy the altitude from the “figures” sheet so that one its endpoints is point P and the other is on the perpendicular you constructed. Label the other endpoint Q. 4. Using point A as center, make a circle with radius AB. 5. Construct a line through point Q perpendicular to PQ (construction marks not shown). Label the point where it intersects the circle point R. 6. Construct radius AR. 7. Using point R as center, mark off a length equal to AB along ray QR. Call the intersection point T. 8. Construct BT. Construct a rhombus given one side and an altitude. Conclusion: ARTB is a rhombus with the required information. Brief Proof: Because AR, AB, and RT, are all radii of congruent circles, they have the same length. Because RT and AB are both perpendicular to PQ, RT is parallel to AB. Therefore, ARTB a parallelogram (one pair of opposite sides are both congruent and parallel). It is now also a rhombus because two adjacent sides are congruent (AR AB). The sides of the rhombus are all congruent to AB, which was the given side. Since QT is parallel to AB, the distance from QT to AB is always the same, and that distance is equal to PQ, the given altitude.