Verified step-by-step solutions for your class notes.
Using a fact about parallelograms, show how to tile the plane using any given triangle.
Answer:
1. Take any given triangle and create an exact duplicate of it.
2. Rotate the duplicate by 180° and align it along one of the sides of the original triangle.
3. The two triangles together form a parallelogram. (Because alternate interior angles are equal, making opposite sides parallel).
4. As discussed at the beginning of the chapter, the plane can always be tiled using copies of any fixed parallelogram by creating a grid of continuous parallel lines. Therefore, by tiling parallelograms, we successfully tile the plane using the given triangle.
You know that the sum of angles of a quadrilateral is 360°, even for a non-convex quadrilateral. Now consider a self-intersecting quadrilateral ABCD, where AB and CD intersect at point E. Show that ∠A + ∠B + ∠C + ∠D < 360°.
Proof:
1. A self-intersecting quadrilateral ABCD with intersecting sides AB and CD at point E forms two distinct triangles: ΔADE and ΔBCE.
2. The sum of the interior angles of any triangle is 180°. So, for ΔADE: ∠A + ∠D + ∠AED = 180°.
3. For ΔBCE: ∠B + ∠C + ∠BEC = 180°.
4. Adding both equations together: ∠A + ∠B + ∠C + ∠D + ∠AED + ∠BEC = 360°.
5. Angles ∠AED and ∠BEC are vertically opposite angles at intersection E, and they have a positive measure greater than 0°.
6. Therefore, ∠A + ∠B + ∠C + ∠D = 360° - (∠AED + ∠BEC). Since we are subtracting a positive value from 360°, it proves that ∠A + ∠B + ∠C + ∠D < 360°.
In a parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ. Show that APCQ is a parallelogram.
Proof:
1. Let the diagonals AC and BD of the original parallelogram ABCD intersect at point O.
2. We know that the diagonals of a parallelogram bisect each other. Therefore, OA = OC and OD = OB.
3. It is given that DP = BQ.
4. Let's find the distances from the center O to points P and Q:
OP = OD - DP
OQ = OB - BQ
5. Since OD = OB and DP = BQ, it must be true that OP = OQ.
6. In the inner quadrilateral APCQ, the diagonals are AC and PQ. We already know OA = OC, and we just proved OP = OQ.
7. Since the diagonals of APCQ bisect each other at O, by the converse of the diagonal property (Theorem 4), APCQ is a parallelogram.
The diagonals AC and BD of a parallelogram ABCD intersect at O. A line through O meets AB and CD at points P and Q respectively. Show that O is the midpoint of PQ.
Proof:
1. Because ABCD is a parallelogram, AB || DC.
2. Consider the triangles ΔAPO and ΔCQO formed by the line passing through O.
3. The diagonals of a parallelogram bisect each other, so OA = OC.
4. Because AB || DC, the alternate interior angles formed by transversal AC are equal: ∠PAO = ∠QCO.
5. The vertically opposite angles at O are equal: ∠AOP = ∠COQ.
6. By Angle-Side-Angle (ASA) congruence, ΔAPO ≅ ΔCQO.
7. Therefore, by Corresponding Parts of Congruent Triangles (CPCTC), OP = OQ. Thus, O is the midpoint of PQ.
Suppose ABCD is a parallelogram and M, N are midpoints of AB and CD respectively. Show that segments DM and BN trisect segment AC.
Proof:
1. ABCD is a parallelogram, so AB || DC and AB = DC.
2. M and N are midpoints, so AM = MB = 1/2 AB, and DN = NC = 1/2 DC.
3. Because AB = DC, it implies AM = NC and MB = DN. Also, MB || DN.
4. A quadrilateral with one pair of equal and parallel opposite sides is a parallelogram (Theorem 5). Thus, MBND is a parallelogram, which means DM || BN.
5. Let DM intersect AC at point P, and BN intersect AC at point Q.
6. In ΔABQ, M is the midpoint of AB, and MP || BQ (since DM || BN). By the Converse of the Midpoint Theorem, P must be the midpoint of AQ. Therefore, AP = PQ.
7. In ΔCDP, N is the midpoint of CD, and NQ || DP (since BN || DM). By the Converse of the Midpoint Theorem, Q must be the midpoint of PC. Therefore, PQ = QC.
8. Combining these gives AP = PQ = QC. The points P and Q divide the diagonal AC into three equal parts, proving that DM and BN trisect AC.
Show that the sum of angles of a non-planar quadrilateral is always less than 360°.
Answer:
Think of a non-planar (skew) quadrilateral ABCD folded along the diagonal AC like a hinge.
1. If the figure were completely flat (planar), the sum of the angles would be exactly 360°.
2. When you fold the quadrilateral along AC to make it non-planar, the points B and D are brought closer together.
3. In 3D space, the straight-line distance between B and D becomes shorter. Because the side lengths AB, BC, CD, DA remain fixed, reducing the distance BD forces the angles at A (∠BAD) and C (∠BCD) to become strictly smaller (by the Law of Cosines for the hinge).
4. The angles at B (∠ABC) and D (∠ADC) remain unchanged because the triangles ΔABC and ΔADC themselves do not warp.
5. Since two interior angles shrink while the other two stay the same, the total sum of the four interior angles must strictly be less than 360°.
Sum of angles of a polygon. What is the sum of angles of a (planar non-self-intersecting) n-gon? Find a formula in terms of n and prove it.
Answer:
1. The sum of the interior angles of a planar n-gon is given by the formula: 180° × (n - 2).
2. Proof: Choose any single vertex of the n-gon. Draw diagonals from this chosen vertex to all other non-adjacent vertices.
3. Since you cannot draw a diagonal to the vertex itself, nor to the two immediately adjacent vertices, you will draw (n - 3) diagonals.
4. These (n - 3) diagonals perfectly divide the interior of the n-gon into exactly (n - 2) distinct triangles.
5. The sum of the interior angles of one triangle is 180°. Because these triangles completely cover the n-gon without overlapping, the sum of all interior angles of the n-gon is simply the sum of the interior angles of these (n - 2) triangles.
6. Therefore, Sum = (n - 2) × 180°.
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