• (i) Adjacent to AB: Sides BC and DA. Opposite to AB: Side CD.
• (ii) Adjacent to ∠A: Angles ∠B and ∠D. Opposite to ∠A: Angle ∠C.
• (iii) Opposite sides are two sides of a quadrilateral that do not share a common vertex. Opposite angles are two internal angles that do not share a common side.

Answer: At any vertex of a quadrilateral, the two adjacent sides form an angle. The internal angle is the angle formed by these two rays that sweeps across the interior region of the quadrilateral. To define it precisely without ambiguity (especially for non-convex cases): It is the angle between the two adjacent sides at the vertex such that a line segment connecting points immediately near the vertex on these two sides lies completely within the quadrilateral's interior. If the quadrilateral is non-convex (has a "dent"), this interior angle will be greater than 180°.

• If AB || DC: No, it cannot be non-convex. If two opposite sides are parallel, the consecutive interior angles must sum to 180° (e.g., ∠A + ∠D = 180°). Since their sum is 180°, neither angle can be greater than 180°. Thus, it must be convex.
• If AB = CD: Yes, it can be non-convex. A self-intersecting "crossed" quadrilateral or a non-convex dart shape can have a pair of opposite sides equal.
• If ∠A = ∠C: Yes, it can be non-convex. A dart shape (non-convex quadrilateral) can be symmetric such that its two non-reflex opposite angles are equal.

Answer: The three lines extending from the triangle ABC divide the plane into 7 regions (the central triangle, and 6 outer regions). Depending on where point D is placed, the resulting sequence of vertices A-B-C-D forms different types of quadrilaterals.

• If D is placed in the regions directly "behind" the vertices (e.g., opposite the triangle from A, B, or C), ABCD will be a self-intersecting (crossed) quadrilateral.
• If D is placed in the regions bounded by the extended sides (the "wedges" pointing outwards from the edges), ABCD will be non-convex (dart-shaped).
• If D is placed such that all internal angles remain less than 180°, it forms a convex quadrilateral.

• (i) True. If one angle is 90°, the opposite angle is also 90°. The adjacent angles must sum to 180°, making them 90° as well. A quadrilateral with four right angles is a rectangle.
• (ii) False. Every rhombus has perpendicular diagonals. It only becomes a square if its diagonals are also equal in length (or if it has a right angle).
• (iii) True. A parallelogram with equal diagonals guarantees that all four angles are 90°, satisfying the definition of a rectangle.

Proof:
1. Since ABCD is a parallelogram, AB || DC and AD || BC.
2. Given AC bisects ∠A, we have ∠DAC = ∠BAC.
3. Because AB || DC, the alternate interior angles are equal: ∠BAC = ∠DCA.
4. Similarly, because AD || BC, the alternate interior angles are equal: ∠DAC = ∠BCA.
5. By substitution, ∠DCA = ∠BCA. Thus, AC also bisects ∠C.
6. In ΔADC, since ∠DAC = ∠DCA, the sides opposite to equal angles are equal: AD = DC.
7. Since adjacent sides of the parallelogram are equal, all four sides must be equal (AB = DC = AD = BC). Therefore, ABCD is a rhombus.

• (i) Yes. If diagonals bisect the angles, you can prove through triangle congruence (ASA) that adjacent sides are equal, making it a rhombus.
• (ii) Yes. If the diagonals bisect each other, it is guaranteed to be a parallelogram. If they do so at right angles, it guarantees all four sides are equal, making it a rhombus.
• (iii) No. An isosceles trapezium also has equal diagonals but is not a rectangle. Extra condition: Add "If the diagonals bisect each other and are of equal length..." then it becomes a rectangle.

Answer:
Let the bisectors of ∠A and ∠B intersect at P, ∠B and ∠C intersect at Q, ∠C and ∠D intersect at R, and ∠D and ∠A intersect at S.
1. In a parallelogram, consecutive angles sum to 180°. Therefore, ∠A + ∠D = 180°.
2. The bisectors of A and D form ΔASD. ∠DAS + ∠ADS = (∠A + ∠D)/2 = 180°/2 = 90°.
3. The third angle of the triangle, ∠ASD, must be 180° - 90° = 90°.
4. By the same logic, the intersections at P, Q, and R also form 90° angles. A quadrilateral PQRS with four right angles is a rectangle.
Why assume AB ≠ BC? If AB = BC, the parallelogram is a rhombus. In a rhombus, the angle bisectors are the diagonals themselves, meaning they intersect at exactly one point rather than forming a new four-sided region.

• (i) By the Midpoint Theorem, PQ || BC, QR || AB, and RP || AC. Because PQ || BR and PR || BQ, PBQR is a parallelogram. The diagonal PR divides it into two congruent triangles: ΔPQR ≅ ΔRBP. By the same logic for the other parallelograms, ΔPQR ≅ ΔQPA ≅ ΔCQR.
• (ii) Yes, you can reconstruct ΔABC. To find the original vertices A, B, and C, simply draw a line through point P parallel to QR, a line through Q parallel to PR, and a line through R parallel to PQ. The intersections of these three lines will perfectly recreate the original ΔABC.

Proof:
1. By the Midpoint Theorem, MN connects the midpoints of AB and AC, therefore MN || BC.
2. Let the segment AD intersect MN at point X.
3. In ΔABD, M is the midpoint of AB, and the segment MX is part of the line MN, which is parallel to the base BD (since MN || BC).
4. By the Converse of the Midpoint Theorem (Theorem 7), a line drawn through the midpoint of one side of a triangle parallel to the base will bisect the third side.
5. Therefore, MX bisects AD, meaning X is the midpoint of AD. Thus, MN bisects AD.

• (i) By Varignon's Theorem (Theorem 9), joining the midpoints of any quadrilateral forms a parallelogram. Thus, PQRS is a parallelogram. In any parallelogram, the diagonals bisect each other. Therefore, PR and QS bisect each other.
• (ii) If the diagonals of the original quadrilateral are equal (AC = BD), then the sides of the Varignon parallelogram PQRS, which are half the lengths of the diagonals (PQ = SR = AC/2 and PS = QR = BD/2), must be equal to each other. A parallelogram with equal adjacent sides is a rhombus. The diagonals of a rhombus are perpendicular. Thus, PR ⊥ QS. Yes, the converse is true: If PR ⊥ QS, then PQRS is a rhombus, meaning its sides are equal, which forces AC = BD.

Answer: Tiling the plane requires shapes to fit together around a vertex such that the sum of the angles meeting at that vertex equals exactly 360°, leaving no gaps or overlaps.
The interior angle of a regular pentagon is 108°.
If we try to fit regular pentagons around a vertex, three pentagons give 108° × 3 = 324°. This leaves a gap of 36°. A fourth pentagon would make the sum 432°, causing an overlap. Because 360 is not evenly divisible by 108, the plane cannot be tiled with regular pentagons.

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.

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°.

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.

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.

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.

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°.

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°.