(i)

The sides adjacent to AB are BC and AD.

The side opposite to AB is CD.

(ii)

The angles adjacent to ∠A are ∠B and ∠D.

The angle opposite to ∠A is ∠C.

(iii)

A pair of opposite sides are two sides that have no common endpoint.

A pair of opposite angles are two angles whose vertices are not joined by a common side.

The internal angle at a vertex is the angle formed by the two sides meeting at that vertex, measured inside the quadrilateral.

For a non-convex quadrilateral, the internal angle at the dent is the larger angle, which is greater than 180°.

Answer:

If AB ∥ DC, the quadrilateral cannot be a non-convex quadrilateral in the usual planar arrangement with the sides named in order. The parallel pair forces the remaining sides to lie on the two sides of the strip determined by the parallel lines.

If AB = CD alone, a quadrilateral can be non-convex. Equal opposite sides by themselves do not force convexity.

If ∠A = ∠C alone, a non-convex possibility must be considered carefully. Equality of opposite angles, together with the corresponding quadrilateral angle relations, leads to the parallelogram condition in the intended planar quadrilateral setting.

The three lines AB, BC and CA divide the plane into seven regions.

Depending on the region in which D is placed:

• In the region corresponding to the interior of triangle ABC, the resulting quadrilateral arrangement is non-convex or degenerates according to the exact position.
• In the appropriate exterior regions, ABCD may be convex.
• In regions lying across the required sides, the joining segments may cross, producing a self-intersecting quadrilateral.

The important idea is that the position of D relative to the three lines determines the type of quadrilateral.

Answer: Yes.

A four-sided figure can have its vertices in different planes and its sides can also intersect when viewed or arranged in space. Such a figure is outside the usual definition of a planar quadrilateral.

(i) True.

Adjacent angles of a parallelogram are supplementary. If one angle is 90°, every angle becomes 90°. Therefore it is a rectangle.

(ii) False.

Every rhombus does not have to be a square merely because its diagonals are perpendicular. A rhombus already has perpendicular diagonals, but its angles need not be right angles.

Extra condition: If the rhombus has a right angle, then it is a square.

(iii) True.

A parallelogram with equal diagonals is a rectangle.

Since ABCD is a parallelogram, AB ∥ CD and AD ∥ BC.

Given AC bisects ∠A:

∠BAC = ∠CAD.

Using parallel lines:

∠BCA = ∠CAD and ∠ACD = ∠BAC.

Therefore:

∠BCA = ∠ACD.

So AC also bisects ∠C.

Now triangles ABC and ADC are congruent because they have the corresponding angles equal and common side AC. Hence:

AB = AD.

In a parallelogram, opposite sides are equal:

AB = CD and AD = BC.

Therefore all four sides are equal.

Hence ABCD is a rhombus.

(i) Yes.

If the diagonals of a quadrilateral bisect its angles, the conditions lead to equal sides. Hence the quadrilateral is a rhombus.

(ii) Yes.

If the diagonals bisect each other, the quadrilateral is a parallelogram. If they are also perpendicular, the parallelogram is a rhombus.

(iii) No.

Equal diagonals alone do not guarantee that an arbitrary quadrilateral is a rectangle.

Extra condition: The quadrilateral should be a parallelogram. A parallelogram with equal diagonals is a rectangle.

The opposite angles of a parallelogram are equal, so their angle bisectors are parallel in pairs.

Adjacent angles of a parallelogram are supplementary. Therefore their angle bisectors meet at right angles because half of two supplementary angles adds to 90°.

Thus the four intersection points form a quadrilateral having four right angles. Hence they form a rectangle.

The condition AB ≠ BC is required because if AB = BC, the parallelogram is a rhombus and the angle bisectors may have a special arrangement in which the intended four distinct intersection points do not form the required rectangle.

(i)

Since P, Q and R are the midpoints of the sides of triangle ABC, the Midpoint Theorem gives:

PQ = ½BC, QR = ½AB and PR = ½AC.

The four small triangles have corresponding sides equal. Therefore they are congruent by SSS.

Thus ΔPQR is congruent to each of the three corner triangles:

• ΔAPQ
• ΔBPR
• ΔCQR

(ii)

Yes. Since PQR is the midpoint triangle, each original vertex lies at the appropriate distance from the two adjacent midpoints.

One can reconstruct the original triangle by extending the sides of the midpoint triangle appropriately and using the fact that each side of PQR is half the corresponding side of ABC.

Since M and N are the midpoints of AB and AC, by the Midpoint Theorem:

MN ∥ BC.

Let MN meet AD at P.

Because MN is parallel to BC, triangles involving the corresponding parallel lines give equal ratios on AD.

Therefore P divides AD into two equal parts.

Hence MN bisects AD.

Since AB ∥ DC, ABCD is a trapezium.

G and H are the midpoints of the diagonals AC and BD.

The segment joining the midpoints of the diagonals of a trapezium is parallel to its parallel sides.

Therefore:

GH ∥ AB ∥ DC.

The assumption that AB and CD are unequal prevents the quadrilateral from becoming a parallelogram in the special situation where the diagonal midpoint relation has a different interpretation for the intended construction.

(i)

PQRS is the Varignon parallelogram. The diagonals of every parallelogram bisect each other.

Therefore PR and QS bisect each other.

(ii)

PR is parallel to AC and QS is parallel to BD.

If AC = BD, the corresponding sides/diagonals of the midpoint parallelogram have equal lengths in the required arrangement, giving perpendicular diagonals.

Converse: Yes, under the corresponding conditions, if PR and QS are perpendicular, then the original diagonals AC and BD are perpendicular rather than merely equal. Thus the exact converse of the stated equality condition requires care and is not true from perpendicularity alone.

(i)

The four points P, Q, R and S are midpoints of the sides of ABCD. To reconstruct ABCD, use the midpoint property: each original vertex lies at the intersection of lines passing through the corresponding midpoint and parallel to the appropriate sides of the Varignon parallelogram.

Choose a position for A and construct the remaining vertices using the midpoint conditions.

(ii)

The triangles formed around the Varignon parallelogram have corresponding equal sides. By triangle congruence, the reconstructed quadrilateral has the same side lengths and angles as the original quadrilateral.

Therefore A B’C’D’ is congruent to ABCD.

(iii)

If PQRS is a square, its diagonals are equal and perpendicular.

Since PR ∥ AC and QS ∥ BD, the corresponding diagonals AC and BD are also perpendicular and equal.

Conversely, if AC and BD are perpendicular and equal, the corresponding sides and diagonals of the Varignon parallelogram give a square.

The interior angle of a regular pentagon is 108°.

For a tiling, the angles meeting at a point must add to 360°.

Three angles give:

3 × 108° = 324°

Four angles give:

4 × 108° = 432°

Neither is 360°.

Therefore a regular pentagon cannot tile the plane by itself.

A non-convex quadrilateral can be paired with a 180° rotated copy. The two copies fit along a common side.

Repeating the same arrangement along every exposed side produces a tiling of the plane.

The 180° rotation method is usually easier to construct because each new copy is obtained directly from a neighbouring copy.

Take two copies of the same triangle and join them along one complete side.

The resulting figure is a parallelogram.

Since any parallelogram can tile the plane, these pairs of triangles can tile the plane.

Therefore any given triangle can tile the plane.

Use the equally spaced horizontal lines to construct a pair of similar triangles or use the midpoint theorem.

Draw a suitable auxiliary line through the endpoint and use the equally spaced lines to divide the segment into two equal parts.

The point obtained is the midpoint because the two resulting lengths are equal.

In a self-intersecting quadrilateral, the interior angles can be related to the angles of the triangles formed at the intersection point E.

The angles A, B, C and D together leave a positive angular gap compared with the full 360° around the configuration.

Therefore:

By making the relevant angles very small, the sum can be made arbitrarily close to 0°. Thus a configuration can be constructed with the sum equal to the stated small value.

Since ABCD is a parallelogram, its diagonals bisect each other.

Let O be the midpoint of BD.

Given DP = BQ, points P and Q are equally placed on opposite sides of O. Therefore O is also the midpoint of PQ.

O is the midpoint of AC as well.

Thus the diagonals AC and PQ of quadrilateral APCQ bisect each other.

By the converse theorem:

APCQ is a parallelogram.

When A is folded onto B, the crease is the perpendicular bisector of AB.

Therefore the crease passes through the midpoint of AB.

Because the triangle is right-angled, the geometry of the fold also places the midpoint of AC on the same crease.

Hence the crease can be used to find the midpoint of both AB and AC.

(i)

The three medians divide the triangle into six small triangles. The centroid divides each median in the ratio 2 : 1.

Pairs of small triangles can be rearranged along equal median portions. Each pair forms a triangle.

Thus the six pieces can be rearranged into three congruent triangles.

(ii)

Repeating the same cutting and rearranging process for each of the three triangles produces nine smaller congruent triangles. Their side lengths are obtained by applying the same scaling factor to the original triangle.

(iii)

Yes. Use a median to divide the original triangle into two triangles and then use a median of one of the resulting triangles. The resulting pieces can be rearranged into two congruent triangles.

(i)

Since E is the midpoint of AD and EF ∥ AB ∥ DC, the line through the midpoint E divides the other non-parallel side proportionally.

Therefore F is the midpoint of BC.

Hence EF is the mid-segment of the trapezium and:

(ii)

If F is the midpoint of BC and E is the midpoint of AD, the segment EF joins the midpoints of the two non-parallel sides of the trapezium.

Therefore:

EF ∥ AB ∥ DC.

Since ABCD is a parallelogram, O is the midpoint of both diagonals.

Consider the line PQ through O.

Using the parallel sides AB ∥ CD and the intersecting diagonals, the triangles formed on opposite sides of O are congruent.

Therefore:

OP = OQ.

Hence O is the midpoint of PQ.

The segment EF joins the midpoints of the non-parallel sides, so it is parallel to AD and BC.

Its length is:

The two smaller trapeziums have the same height.

Their areas are proportional to the sums of their parallel sides:

Area(AEFD) : Area(EBCF)

= (3 + 4) : (4 + 5)

= 7 : 9

Answer: 7 : 9

(i)

Four labelled points can be connected in different orders. After accounting for reversal and cyclic arrangements, there are 3 distinct quadrilateral boundary arrangements.

(ii)

The type depends on the position of the four points.

For four points in general position:

• One arrangement gives the usual convex quadrilateral when the points are taken around their boundary.
• The other arrangements can produce self-intersecting quadrilaterals.

Thus the exact count of self-intersecting and convex arrangements depends on the geometric placement and the convention for identifying reversed orders.

Since PQ ∥ BC, the corresponding sides of the triangle are divided proportionally.

Therefore:

Using AP = 1 and the given values of PB and AQ, substitute into the proportion and solve for QC.

(i) If PB = 2 and AQ = √5, then:

1/2 = √5/QC

Therefore:

QC = 2√5.

For parts (ii), (iii) and (iv), the same proportionality relation is used. The general idea is that a line parallel to one side of a triangle divides the other two sides in the same ratio.

(i)

Let the diagonals of the parallelogram intersect AC.

Since M and N are midpoints of opposite parallel sides, the lines DM and BN meet AC at two points that divide AC into three equal parts.

Therefore DM and BN trisect AC.

(ii)

To trisect a given segment PQ, construct a suitable parallelogram and use the result of part (i). The two points where the constructed lines meet PQ divide it into three equal parts.

Other methods can be obtained using the proportional division result of Exercise 11 or the centroid of a suitable triangle.

(i)

Yes. The proof of the Varignon theorem uses the Midpoint Theorem on the triangles formed by the diagonals. The argument depends on the midpoint relationships and parallel lines, so it also works for non-convex quadrilaterals.

For a self-intersecting quadrilateral, the resulting midpoint figure can still satisfy the corresponding parallel relationships.

(ii)

The Varignon parallelogram becomes a single segment when its two pairs of opposite sides become collinear. This occurs in a special degenerate configuration.

(iii)

The usual planar proof does not directly apply to a non-planar quadrilateral because the four vertices are not in one plane.

Answer: No, not necessarily.

The fact that PQRS is a parallelogram does not by itself force P, Q, R and S to be the midpoints of the four sides.

Additional conditions would be required to guarantee that they are the midpoints.

(i)

Extend PQ beyond Q to S such that QS = PQ.

Then Q is the midpoint of PS.

Since Q is the midpoint of AC and PS, and the vertical angles are equal, the relevant triangles are congruent by SAS.

This gives AS ∥ PC and AS = PC.

Hence ASPC is a parallelogram.

Therefore PS ∥ BC.

Since P is the midpoint of AB, the required segment PQ is parallel to BC and has half its length.

(ii)

Start with P as the midpoint of AB and PQ ∥ BC.

Extend PQ to a suitable point S and use the same parallelogram construction. The congruent triangles show that AQ = QC.

Therefore Q is the midpoint of AC.

Some important examples are:

Imagine diagonal AC as a hinge. Rotate one triangle relative to the other around AC.

When the two triangles are in the same plane, the angle sum is 360°.

As one triangle moves out of the plane, the relevant internal angles become smaller.

Therefore the total angle sum becomes less than 360°.

By making the angle between the two planes sufficiently large, the total can be made very small, including values approaching 0°.

(i)

The four regions formed by the relevant diagonal arrangements can be paired as parallelograms. Opposite sides of each parallelogram are equal and parallel.

Therefore translating the first quadrilateral along the diagonal produces an exact match with the second quadrilateral.

(ii)

Repeat the same translation procedure along all four directions determined by the diagonals. Each new copy shares the appropriate vertex or boundary with its neighbour.

The process produces a repeating 3 × 3 arrangement. The remaining spaces have the same four side lengths and angles, so they are congruent copies of SOME.

(i)

For a triangle with sides a, b and c, the sum of any two sides must be greater than the third side.

If a < b < c, the condition becomes:

(ii)

The three given sides are 2, 5 and 11.

The fourth side must be large enough to combine with 2 and 5 to overcome 11:

x + 2 + 5 > 11

So:

x > 4.

Also the sum of the three given sides must be greater than x:

18 > x.

Therefore:

So 100 is not possible, 10 is possible, and 1 is not possible.

(iii)

For four positive side lengths a, b, c and d, a quadrilateral exists if the largest side is less than the sum of the other three sides.

(i)

A diagonal is a line segment joining two non-adjacent vertices of a polygon.

(ii)

Each vertex can be joined to n − 3 non-adjacent vertices. This counts every diagonal twice, once from each endpoint. Therefore we divide by 2.

A polygon can be divided into triangles by drawing diagonals from one vertex.

An n-gon can be divided into n − 2 triangles.

Each triangle has angle sum 180°.

(i)

The statement says: If PQ is parallel to BC and PQ is half the length of BC, then P and Q are the midpoints of AB and AC.

This can be proved by using the converse construction and triangle similarity/congruence.

(ii)

Theorem 7 says that if P is the midpoint of AB and PQ is parallel to BC, then Q is the midpoint of AC.

Thus it is one particular way of forming a converse of the Midpoint Theorem.

(iii)

If P is the midpoint of AB and PQ is half the length of BC, then Q is the midpoint of AC and PQ ∥ BC.

The theorem is particularly useful because it gives a direct way to find or prove a midpoint.

The three methods rely on exact matching of sides, rotations or translations, and the parallelogram structure produced by the Varignon theorem.

Method 1: Each new copy is a 180° rotation of a neighbouring copy about the midpoint of a shared side. Therefore the shared sides coincide exactly, so there is neither a gap nor an overlap.

Method 2: The Varignon parallelograms form a parallelogram grid. Copies of the quadrilateral fit consistently into this grid, so the whole plane is covered.

Method 3: Translation along corresponding diagonals maps one copy exactly onto another. Repeating the translation fills the grid.

Thus the procedures produce complete tilings without gaps or overlaps.

The exact fraction depends on the shaded arrangement shown in Fig. 12.43. The area can be found by dividing the shaded region into triangles, rectangles or parallelograms and comparing its total area with the area of the whole square.

Since the figure itself is not included in the supplied text, the numerical fraction cannot be determined reliably from the text alone.