NCERT Solutions for Class 9 Maths Chapter 12 — End-of-Chapter Exercises#

Take the triangle $ABC$ and a copy turned through 180° about the midpoint $M$ of side $BC$. The half-turn sends $B$ to $C$ and $C$ to $B$, and sends $A$ to a new point $D$ with $M$ the midpoint of $AD$.

The quadrilateral $ABDC$ has diagonals $AD$ and $BC$ that bisect each other at $M$, so it is a parallelogram (Theorem 4).

Parallelograms tile the plane in a grid (the introduction). Each parallelogram is two copies of the triangle, so the triangle tiles the plane.

In Fig. 12.33 the segment starts on the 2nd line and ends on the 6th line (counting from the top), crossing the 3rd, 4th and 5th lines. Its midpoint is the point where it crosses the 4th line — two gaps from each end.

Justification: equally spaced parallel lines cut any line crossing them into equal parts. Take any three consecutive lines; the middle one is halfway between the outer two. By Exercise 7(i) below (a line through the midpoint of one side of a trapezium, parallel to the parallel sides, bisects the other side) — or by Theorem 7 when the segment starts at a crossing — the middle line bisects the part of the segment between the outer two. So the four pieces between lines 2 and 6 are equal, and the crossing with line 4 is the midpoint.

The sides $AB$ and $CD$ cross at $E$, making two triangles: $\triangle EBC$ (with angles $\angle B$ and $\angle C$ of the quadrilateral) and $\triangle EAD$ (with $\angle A$ and $\angle D$).

$$\angle B + \angle C = 180° - \angle BEC$$

$$\angle A + \angle D = 180° - \angle AED$$

$\angle BEC = \angle AED$ (vertically opposite angles). Adding:

$$\angle A + \angle B + \angle C + \angle D = 360° - 2\angle BEC$$

$\angle BEC > 0°$, so the sum is less than 360°.

We need $360° - 2\angle BEC = 2°$, that is, $\angle BEC = 179°$.

Construction: draw two lines through a point $E$ that meet at 179° (they look almost like one straight line). Mark $B$ and $C$ on the two lines on one side so that $\angle BEC = 179°$, and $A$ and $D$ on the extensions beyond $E$ (so $A$, $E$, $B$ are collinear and $C$, $E$, $D$ are collinear). Then $ABCD$ (sides $AB$, $BC$, $CD$, $DA$) is self-intersecting at $E$, and

$$\angle A + \angle B + \angle C + \angle D = 360° - 2 \times 179° = 2°$$

(Both triangles $EBC$ and $EAD$ are extremely thin.)

Let the diagonals $AC$ and $BD$ meet at $O$. The diagonals of a parallelogram bisect each other:

$$OA = OC, \quad OB = OD$$

$P$ is between $D$ and $O$, and $Q$ is between $O$ and $B$, so:

$$OP = OD - DP$$

$$OQ = OB - BQ$$

Since $OD = OB$ and $DP = BQ$:

$$\Rightarrow OP = OQ$$

So in quadrilateral $APCQ$, the diagonals $AC$ and $PQ$ bisect each other at $O$. By Theorem 4, $APCQ$ is a parallelogram.

In the figure, the right angle is at $B$, so $AB \perp BC$.

Folding $A$ onto $B$ makes the crease the perpendicular bisector of $AB$: it passes through the midpoint $M$ of $AB$ and is perpendicular to $AB$.

Both the crease and $BC$ are perpendicular to $AB$, so the crease is parallel to $BC$.

In $\triangle ABC$, the crease is a line through the midpoint $M$ of $AB$, parallel to $BC$. By Theorem 7, it bisects the third side $AC$.

∴ The crease meets $AB$ at its midpoint and $AC$ at its midpoint.

Here $P$, $Q$, $R$ are the midpoints of $BC$, $CA$, $AB$. By the Centroid Theorem, $AM = 2MP$, $BM = 2MQ$, $CM = 2MR$.

Assembling one triangle: take $\triangle MPB$ and $\triangle MPC$. Since $PB = PC$, turn $\triangle MPB$ over and place it against $\triangle MPC$ so that $P$ matches $P$ and $B$ matches $C$. Let $M'$ be the new position of $M$.

The angles at $P$ fit together:

$$\angle MPC + \angle M'PC = \angle MPC + \angle MPB$$

$$\angle MPC + \angle M'PC = 180°$$

So $M$, $P$, $M'$ lie on a straight line, and the two pieces form a triangle $MCM'$ with sides:

$$MM' = 2MP = AM, \quad MC = CM, \quad M'C = MB = BM$$

In the same way, $\triangle MQC$ and $\triangle MQA$ (with $QC = QA$) form a triangle with sides $2MQ = BM$, $CM$, $AM$; and $\triangle MRA$ and $\triangle MRB$ form a triangle with sides $2MR = CM$, $AM$, $BM$.

All three triangles have sides $AM$, $BM$, $CM$, so they are congruent (SSS). Their side lengths are

$$AM = \dfrac{2}{3}AP, \quad BM = \dfrac{2}{3}BQ, \quad CM = \dfrac{2}{3}CR$$

— two-thirds of the medians of $\triangle ABC$.

Call the assembled triangle $T$. In the triangle $MCM'$ from (i), $P$ is the midpoint of $MM'$, so $CP$ is a median of $T$, of length $\dfrac{BC}{2}$, drawn to the side of length $AM$.

The other two assembled triangles are congruent copies of $T$, and in them $Q$ and $R$ play the same role: $T$ has a median of length $QC = \dfrac{AC}{2}$ to the side of length $BM$, and a median of length $RA = \dfrac{AB}{2}$ to the side of length $CM$.

So the three medians of $T$ are $\dfrac{BC}{2}$, $\dfrac{AC}{2}$, $\dfrac{AB}{2}$. Applying (i) to $T$, each new triangle has sides equal to $\dfrac{2}{3}$ of these medians:

$$\dfrac{2}{3} \times \dfrac{BC}{2} = \dfrac{BC}{3}, \quad \dfrac{2}{3} \times \dfrac{AC}{2} = \dfrac{AC}{3}, \quad \dfrac{2}{3} \times \dfrac{AB}{2} = \dfrac{AB}{3}$$

∴ All 9 triangles have sides $\dfrac{AB}{3}$, $\dfrac{BC}{3}$, $\dfrac{AC}{3}$ — each is a one-third-size copy of $\triangle ABC$.

1. Cut along the median $AP$ ($P$ the midpoint of $BC$). This gives $\triangle APB$ and $\triangle APC$.
2. In $\triangle APB$, let $R$ be the midpoint of $AB$, and cut along the median $PR$. This gives $\triangle PRA$ and $\triangle PRB$.
3. Turn $\triangle PRB$ through 180° about $R$. Then $B$ goes to $A$ (as $R$ is the midpoint of $AB$), and $P$ goes to a point $P'$ on line $PR$ with $RP' = RP$.

The two pieces now form triangle $APP'$ with sides:

$$AP, \quad AP' = BP = PC, \quad PP' = 2PR = AC$$

(using the Midpoint Theorem in $\triangle ABC$: $PR = \dfrac{AC}{2}$).

These are exactly the sides of $\triangle APC$: $AP$, $PC$, $CA$. So $\triangle APP' \cong \triangle APC$ (SSS), and the triangle has been cut into pieces forming 2 congruent triangles.

Join $AC$, and let $EF$ meet $AC$ at $G$.

In $\triangle ADC$: $E$ is the midpoint of $AD$ and $EG \parallel DC$. By Theorem 7, $G$ is the midpoint of $AC$, and

$$EG = \dfrac{1}{2}DC$$

In $\triangle CAB$: $G$ is the midpoint of $CA$ and $GF \parallel AB$. By Theorem 7, $F$ is the midpoint of $CB$, and

$$GF = \dfrac{1}{2}AB$$

Adding:

$$EF = EG + GF$$

$$EF = \dfrac{1}{2}(AB + CD)$$

Method 1 (direct): let $M$ be the midpoint of $BD$. In $\triangle DAB$, $E$ and $M$ are midpoints, so $EM \parallel AB$. In $\triangle BDC$, $M$ and $F$ are midpoints, so $MF \parallel DC \parallel AB$. Both $EM$ and $MF$ pass through $M$ and are parallel to $AB$, so they are the same line. So $E$, $M$, $F$ are collinear and $EF \parallel AB$.

Method 2 (using (i)): draw the line through $E$ parallel to $AB$. By (i), it meets $BC$ at its midpoint. But $F$ is the midpoint of $BC$, so this line passes through $F$ — it is the line $EF$. So $EF \parallel AB$.

Method 2 is shorter, because part (i) has already done the work.

In $\triangle OAP$ and $\triangle OCQ$:

Part	Reason
$OA = OC$	diagonals of a parallelogram bisect each other
$\angle AOP = \angle COQ$	vertically opposite angles
$\angle OAP = \angle OCQ$	alternate angles ($AB \parallel DC$, transversal $AC$)

By ASA, $\triangle OAP \cong \triangle OCQ$, so $OP = OQ$.

∴ $O$ is the midpoint of $PQ$. This congruence proof is probably the simplest.

By Exercise 7, $EF$ is parallel to the parallel sides, and

$$EF = \dfrac{3 + 5}{2} = 4 \text{ cm}$$

Since $E$ is the midpoint of $AB$, the line $EF$ is halfway between $AD$ and $BC$. If the trapezium’s height is $h$, each part has height $\dfrac{h}{2}$:

$$\text{Area}(AEFD) = \dfrac{1}{2}(3 + 4) \times \dfrac{h}{2} = \dfrac{7h}{4}$$

$$\text{Area}(EBCF) = \dfrac{1}{2}(4 + 5) \times \dfrac{h}{2} = \dfrac{9h}{4}$$

∴ The ratio of the areas is 7 : 9.

A quadrilateral is fixed by the order in which we go round the four points. Starting at $A$ and ignoring direction, there are only 3 different orders: $ABCD$, $ABDC$, $ACBD$. So the points form 3 quadrilaterals.

It depends on how the points lie:

Arrangement	Convex	Non-convex	Self-intersecting
No point inside the triangle of the other three	1	0	2
One point inside the triangle of the other three	0	3	0

When the four points are like the corners of a convex shape, going round in order gives the convex quadrilateral, and the other two orders cross. When one point is inside the triangle of the other three, all three quadrilaterals have a dent at that point and none crosses.

$PB = 2$ is two steps of length 1. Mark the point on $PB$ that is 1 from $P$, and draw the dotted line through it parallel to $BC$. The parallels through $A$, $P$, this point and $B$ are equally spaced along $AB$, so (Exercise 2 above) they cut $AC$ into equal parts, each of length $AQ = \sqrt{5}$.

$$QC = 2\sqrt{5}$$

Divide $AP = 1$ into three parts of $\dfrac{1}{3}$ (the dotted lines). Now $AB$ is cut into 4 equal parts of $\dfrac{1}{3}$ ($AP$ has 3, $PB$ has 1), and the parallels through these points cut $AC$ into 4 equal parts. $AQ = \sqrt{5}$ is 3 of them, so each part is $\dfrac{\sqrt{5}}{3}$:

$$QC = \dfrac{\sqrt{5}}{3}$$

Divide both $AP$ and $PB$ into parts of $\dfrac{1}{3}$: $AP$ has 3 parts and $PB$ has 2. The parallels cut $AC$ into 5 equal parts, 3 of which make $AQ = \sqrt{5}$:

$$\text{One part} = \dfrac{\sqrt{5}}{3}$$

$$QC = 2 \times \dfrac{\sqrt{5}}{3} = \dfrac{2\sqrt{5}}{3}$$

Let $\dfrac{AP}{PB} = \dfrac{m}{n}$ with $m$, $n$ natural numbers. Then $AP = mu$ and $PB = nu$ for some length $u$. Divide $AB$ into $m + n$ equal parts of length $u$, so that $P$ is one of the division points, and draw lines parallel to $BC$ through all of them.

These lines are equally spaced, so they cut $AC$ into $m + n$ equal parts of some length $v$ (each middle line bisects the part of $AC$ between its neighbours, by Exercise 7(i)). The line through $P$ is the line $PQ$, so:

$$AQ = mv, \quad QC = nv$$

$$\Rightarrow \dfrac{AQ}{QC} = \dfrac{m}{n} = \dfrac{AP}{PB}$$

$MB = \dfrac{1}{2}AB = \dfrac{1}{2}CD = DN$, and $MB \parallel DN$. By Theorem 5, $MBND$ is a parallelogram, so $DM \parallel BN$.

Let $DM$ meet $AC$ at $X$, and $BN$ meet $AC$ at $Y$.

In $\triangle ABY$: $M$ is the midpoint of $AB$ and $MX \parallel BY$. By Theorem 7, $X$ is the midpoint of $AY$, so $AX = XY$.

In $\triangle CDX$: $N$ is the midpoint of $CD$ and $NY \parallel DX$. By Theorem 7, $Y$ is the midpoint of $CX$, so $XY = YC$.

∴ $AX = XY = YC$: $DM$ and $BN$ trisect $AC$.

Way 1 (part (i)): take $PQ$ as the diagonal $AC$. Choose any point $B$ not on $PQ$, and complete the parallelogram $PBQD$ (draw the lines through $P$ and $Q$ parallel to $BQ$ and $BP$). Mark the midpoints $M$ of $PB$ and $N$ of $QD$. The lines $DM$ and $BN$ cut $PQ$ into three equal parts.

Way 2 (Exercise 11): from $P$ draw any ray, and mark three equal steps $P X_1 X_2 X_3$ on it. Join $X_3Q$. Draw lines through $X_1$ and $X_2$ parallel to $X_3Q$. They meet $PQ$ at its trisection points.

Way 3 (Centroid Theorem): make $PQ$ a median of a triangle: choose any point $B$, and let $C$ be the point with $Q$ as the midpoint of $BC$. Then $PQ$ is the median from $P$ in $\triangle PBC$. Draw a second median (from $B$ to the midpoint of $PC$); it meets $PQ$ at the centroid $G$, with $PG = 2GQ$, so $G$ is one trisection point. The midpoint of $PG$ is the other.

The proof uses only the Midpoint Theorem in the four triangles $ABC$, $ADC$, $BCD$, $BAD$:

$$PQ \parallel AC \parallel SR, \quad PQ = SR = \dfrac{1}{2}AC$$

$$QR \parallel BD \parallel PS, \quad QR = PS = \dfrac{1}{2}BD$$

These triangles exist whatever the shape of $ABCD$ — convex, non-convex or self-intersecting — so the conclusion still holds.

$PQ$ is parallel to $AC$ and $QR$ is parallel to $BD$. If the lines $AC$ and $BD$ are parallel (or the same line), then $PQ$ and $QR$ lie along one line, and all four midpoints are collinear — the “parallelogram” flattens into a segment.

This happens for a self-intersecting $ABCD$ whose “diagonals” $AC$ and $BD$ are parallel — for example, when $A$, $C$ lie on one line and $B$, $D$ on a parallel line, with $ABCD$ crossing over (like a bow-tie made from an isosceles trapezium).

Yes. Each of the four triangles is planar, so the Midpoint Theorem applies in each. Then $PQ$ and $SR$ are both parallel to $AC$ and equal to $\dfrac{1}{2}AC$. Two parallel segments lie in one plane, so $P$, $Q$, $R$, $S$ are coplanar, and $PQRS$ has a pair of opposite sides equal and parallel — it is a parallelogram (Theorem 5). The Varignon parallelogram is flat even when $ABCD$ is not!

No. Take a square $ABCD$ of side 4, and mark $P$, $Q$, $R$, $S$ at a distance 1 from $A$, $B$, $C$, $D$ along each side (going round the same way). By symmetry under a quarter-turn about the centre, $PQRS$ is a square — certainly a parallelogram — but none of the points is a midpoint (each side is cut 1 : 3).

$P$ and $Q$ are the midpoints of $AB$ and $AC$. Extend $PQ$ beyond $Q$ by its own length, to $S$, so that $QS = PQ$.

In $\triangle APQ$ and $\triangle CSQ$: $AQ = CQ$, $\angle AQP = \angle CQS$ (vertically opposite), and $PQ = SQ$. By SAS, $\triangle APQ \cong \triangle CSQ$. So:

• $CS = AP = PB$;
• $\angle QCS = \angle QAP$, which are alternate angles for lines $CS$ and $AB$ with transversal $AC$, so $CS \parallel AB$, that is, $CS \parallel PB$.

$PBCS$ has a pair of opposite sides ($PB$ and $CS$) equal and parallel, so it is a parallelogram (Theorem 5). Hence:

$$PS \parallel BC \quad \text{and} \quad PS = BC$$

$$\Rightarrow PQ \parallel BC, \quad PQ = \dfrac{1}{2}PS = \dfrac{1}{2}BC$$

$P$ is the midpoint of $AB$, and the line through $P$ parallel to $BC$ meets $AC$ at $Q$. Extend $PQ$ to the point $S$ where it meets the line through $C$ parallel to $AB$.

$PBCS$ has both pairs of opposite sides parallel, so it is a parallelogram, and $CS = PB = PA$.

In $\triangle APQ$ and $\triangle CSQ$: $AP = CS$, $\angle PAQ = \angle SCQ$ (alternate angles, $AB \parallel CS$), and $\angle AQP = \angle CQS$ (vertically opposite). By AAS, $\triangle APQ \cong \triangle CSQ$, so $AQ = CQ$: $Q$ is the midpoint of $AC$. Also $PQ = QS = \dfrac{1}{2}PS = \dfrac{1}{2}BC$.

Property	Converse	True?
Diagonals of a rhombus are perpendicular bisectors of each other	A quadrilateral whose diagonals are perpendicular bisectors of each other is a rhombus	Yes (Ex. Set 12.2, Q3(ii))
Diagonals of a rhombus bisect its angles	A quadrilateral whose diagonals bisect its angles is a rhombus	Yes (Ex. Set 12.2, Q3(i))
Diagonals of a rhombus are perpendicular	A quadrilateral with perpendicular diagonals is a rhombus	No — a kite
Diagonals of a rectangle are equal and bisect each other	A quadrilateral whose diagonals are equal and bisect each other is a rectangle	Yes
Diagonals of a rectangle are equal	A quadrilateral with equal diagonals is a rectangle	No — an isosceles trapezium
Diagonals of a square are equal, perpendicular and bisect each other	A quadrilateral with such diagonals is a square	Yes
All sides of a square are equal	A quadrilateral with all sides equal is a square	No — a rhombus
All angles of a square are equal	A quadrilateral with all angles equal is a square	No — a rectangle

Split $ABCD$ along the diagonal $AC$ into $\triangle ABC$ and $\triangle ADC$. Their six angles add up to $180° + 180° = 360°$.

$\angle B$ and $\angle D$ are angles of these triangles. But $\angle A$ (between $AB$ and $AD$) and $\angle C$ are not: when the two triangles do not lie in one plane, the angle between $AB$ and $AD$ is less than $\angle BAC + \angle CAD$ (the “bend” at the hinge makes the direct angle smaller), and similarly at $C$. So:

$$\angle A + \angle B + \angle C + \angle D < 360°$$

A sum of 2°: take $\triangle ABC$ and $\triangle ADC$ very thin, with $\angle B = \angle D = 0.5°$ (so $B$ and $D$ are far from $AC$). Now rotate $\triangle ADC$ about the hinge $AC$ until it nearly lies on top of $\triangle ABC$. Then $AD$ nearly lines up with $AB$, so $\angle A$ shrinks towards 0°, and likewise $\angle C$. $\angle B$ and $\angle D$ stay 0.5° each. Stopping when $\angle A = \angle C = 0.5°$ gives

$$\angle A + \angle B + \angle C + \angle D = 2°$$

The slide moves every point of SOME the same distance in the same direction — the distance $EO$ along the line $EO$. For each vertex $X$ of SOME and its image $X'$, the segments $XX'$ are therefore all equal and parallel to $EO$.

Taking the vertices two at a time, the four quadrilaterals $SS'O'O$, $OO'M'M$, $MM'E'E$ and $EE'S'S$ each have a pair of opposite sides ($XX'$ and $YY'$) equal and parallel, so each is a parallelogram (Theorem 5). In a parallelogram the other pair of sides is also equal and parallel, so every side of SOME matches the corresponding side of $S'O'M'E'$: $SO = S'O'$, $OM = O'M'$, $ME = M'E'$, $ES = E'S'$, each parallel to its partner. The two 4-gons coincide exactly.

Sliding the starting copy along the diagonal $EO$ in both directions, and along the diagonal $SM$ in both directions, gives 4 new copies, each meeting the first only at a vertex. Repeating the same slides from these gives the corners of a $3 \times 3$ grid of 9 copies. The gaps between them have sides that are copies of the sides of SOME and angles that are copies of its angles, so each gap is also a copy of SOME, completing the tiling.

Draw the longest side $c$. From its ends draw arcs of radius $a$ and $b$. They meet (in a point off the line) exactly when $a + b > c$; if $a + b \leq c$ the arcs do not meet off the line, and no triangle forms.

A 4-gon can be formed exactly when each side is less than the sum of the other three (the longest side is the only one that needs checking).

• $x = 100$: $100 < 2 + 5 + 11 = 18$? No → impossible.
• $x = 10$: the longest is 11, and $11 < 2 + 5 + 10 = 17$ ✓ → possible.
• $x = 1$: $11 < 2 + 5 + 1 = 8$? No → impossible.

In general we need $x < 18$ and $11 < 7 + x$:

$$4 < x < 18$$

There is a 4-gon with sides $a$, $b$, $c$, $d$ exactly when the longest of them is less than the sum of the other three. (Then we can hinge the sides open like a linkage until the shape closes up with no three vertices collinear.)

A diagonal of an $n$-gon is a segment joining two vertices that are not adjacent (not joined by a side).

$n$	3	4	5	6	7	8
Diagonals	0	2	5	9	14	20

$$\text{Diagonals} = \dfrac{n(n - 3)}{2}$$

Increase by one side: going from $n$ to $n + 1$ sides adds a new vertex. It can be joined to $(n + 1) - 3 = n - 2$ non-adjacent vertices, and the old side that the new vertex “breaks” becomes a diagonal, so $(n - 2) + 1 = n - 1$ diagonals are added (e.g. 5 → 9 adds 4 when going from 5 to 6 sides).

Justification: each vertex can be joined to $n - 3$ others (not itself, and not its two neighbours). That gives $n(n - 3)$ ends of diagonals, but every diagonal has two ends, so it was counted twice:

$$\text{Diagonals} = \dfrac{n(n - 3)}{2}$$

$n$	3	4	5	6	7
Sum of angles	180°	360°	540°	720°	900°

$$\text{Sum of angles} = (n - 2) \times 180°$$

Proof: an $n$-gon can be cut by $n - 3$ diagonals into $n - 2$ triangles, with the angles of the triangles together making up exactly the angles of the polygon. For a convex polygon, draw all diagonals from one vertex. For a non-convex polygon, there is always some diagonal lying inside it; cutting along it gives two smaller polygons, and we keep cutting until only triangles remain. Each cut increases the number of pieces by 1, so $n - 3$ cuts give $n - 2$ triangles, whose angles total $(n - 2) \times 180°$.

Statement: In $\triangle ABC$, let $P$ be on side $AB$ and $Q$ on side $AC$. If $PQ \parallel BC$ and $PQ = \dfrac{1}{2}BC$, then $P$ and $Q$ are the midpoints of $AB$ and $AC$.

Proof: through $Q$, draw the line parallel to $AB$, meeting $BC$ at $X$. Then $PQXB$ has both pairs of opposite sides parallel, so it is a parallelogram, and

$$BX = PQ = \dfrac{1}{2}BC$$

So $X$ is the midpoint of $BC$. In $\triangle CBA$, the line $XQ$ passes through the midpoint $X$ of $CB$ and is parallel to $BA$, so by Theorem 7 it bisects $CA$: $Q$ is the midpoint of $AC$. Then the line $QP$ through the midpoint $Q$ of $AC$, parallel to $CB$, bisects $AB$: $P$ is the midpoint of $AB$.

Write the Midpoint Theorem as: “Suppose $P$ is the midpoint of side $AB$ of $\triangle ABC$ and $Q$ is a point on side $AC$. If $Q$ is the midpoint of $AC$, then $PQ \parallel BC$.”

Keeping the first sentence and reversing the second gives: “Suppose $P$ is the midpoint of side $AB$ of $\triangle ABC$ and $Q$ is a point on side $AC$. If $PQ \parallel BC$, then $Q$ is the midpoint of $AC$.” This is exactly Theorem 7.

In terms of the named conditions: if (P MID) and (PRLL), then (Q MID) and (HALF).

The pairs of conditions we have used as assumptions are: (P MID)+(Q MID) (the theorem), (PRLL)+(HALF) (part (i)), and (P MID)+(PRLL) or (Q MID)+(PRLL) (Theorem 7). The only new combination is (P MID) with (HALF) (or the same with $Q$).

Question: “Suppose $P$ is the midpoint of $AB$, $Q$ is on $AC$, and $PQ = \dfrac{1}{2}BC$. Must $Q$ be the midpoint of $AC$, and must $PQ \parallel BC$?”

Answer: No. A circle with centre $P$ and radius $\dfrac{1}{2}BC$ passes through the midpoint $N$ of $AC$, but it can cut $AC$ at a second point. Example: $A(0, 0)$, $B(4, 0)$, $C(4, 2)$, so $P = (2, 0)$ and $\dfrac{1}{2}BC = 1$. The point $Q = \left(\dfrac{6}{5}, \dfrac{3}{5}\right)$ is on $AC$, and

$$PQ^2 = \left(2 - \dfrac{6}{5}\right)^2 + \left(\dfrac{3}{5}\right)^2 = \dfrac{16}{25} + \dfrac{9}{25} = 1$$

so $PQ = \dfrac{1}{2}BC$, but $Q$ is not the midpoint $(2, 1)$ of $AC$, and $PQ$ is not parallel to $BC$.

To prove a procedure gives a tiling we must show two things: the copies never overlap, and they leave no gaps anywhere in the plane.

• Method 1 (half-turns about midpoints): two copies sharing a side, related by a half-turn about its midpoint, lie on opposite sides of that side, so they do not overlap. Around each vertex, the four copies bring the four different angles of the 4-gon, which add up to 360°, so they close up exactly with no gap. Because each new copy is obtained from either neighbour by the same half-turn, the copies agree with each other everywhere, and the pattern extends without conflict.
• Method 2 (Varignon grid): the Varignon parallelograms tile the plane. Each 4-gon is made of its own parallelogram plus four corner triangles, and each corner triangle exactly fills half of a neighbouring parallelogram of the grid. Since the parallelograms cover the plane with no gaps or overlaps, so do the 4-gons.
• Method 3 (slides along diagonals): slides do not change shape, and the copies placed by sliding along the two diagonals form a repeating grid pattern. The gaps between copies have the same sides and angles as the 4-gon, so they are copies too, and the whole plane is covered.

Each line joins a corner of the square to the midpoint of a side, as in Chapter 6 (Fig. 6.44). Place the square with corners $(0, 0)$, $(2, 0)$, $(2, 2)$, $(0, 2)$, so its area is 4. The four lines come in two pairs of parallel lines, each pair $\dfrac{2}{\sqrt{5}}$ apart, and the two pairs are perpendicular, so the shaded region is a square of side $\dfrac{2}{\sqrt{5}}$:

$$\text{Shaded} = \left(\dfrac{2}{\sqrt{5}}\right)^2 = \dfrac{4}{5}$$

$$\dfrac{\text{Shaded}}{\text{Square}} = \dfrac{4/5}{4} = \dfrac{1}{5}$$

∴ One-fifth of the square is shaded.