Quadrilaterals

1. (i) Sides adjacent to AB: AD, BC (each shares an endpoint with AB). Side opposite AB: CD (shares no endpoint with AB).
2. (ii) Angles adjacent to ∠A: ∠B and ∠D. Angle opposite ∠A: ∠C.
3. (iii) Opposite sides are two sides that have no common vertex. Opposite angles are angles at vertices that are not endpoints of a common side (equivalently, their vertices are joined by a diagonal).

Imagine drawing a tiny circle centred at A, small enough to miss the two sides that do not meet there. The rays AB and AD cut the circle into two arcs. The interior angle at A is the angle swept out along the arc inside the quadrilateral. At an outward corner this angle is less than 180°. At the inward corner of a concave quadrilateral it is greater than 180°.

The direction of the opposite vertex alone can be misleading. For example, if B is the dent, the ray AC can lie outside the interior angle at A.

AB ∥ DC: No. A simple quadrilateral with opposite parallel sides is a convex trapezium (a parallelogram is a special case). Trying to make it concave would force the side segments to cross.

AB = CD: Yes. Equal lengths alone do not imply parallelism or convexity. For example A=(0,0), B=(3,0), C=(1,0.5), D=(1,3.5) has AB=CD=3, but C is a reflex vertex.

∠A = ∠C: Yes. A symmetric dart with a dent at B or D can have equal opposite angles A and C. Equality of both pairs of opposite angles, not just one pair, forces a parallelogram.

Take A, B, C in counter-clockwise order. Describe D by which of the three side-lines it crosses when compared with the interior of △ABC:

Why is the case across CA special? With ABCD listed in that order, placing D beyond CA creates the natural convex four-corner boundary. Crossing only AB or BC makes a pair of opposite sides intersect.

No, not with straight sides. If opposite sides AB and CD intersect at some point E, then the two intersecting lines AB and CD lie in a single plane. All four endpoints A, B, C and D lie in that same plane. Thus the figure cannot be non-planar.

1. (i) True. A parallelogram has adjacent supplementary angles. If one is 90°, all four are 90°.
2. (ii) False. All rhombuses already have perpendicular diagonals. A non-square rhombus is an immediate counterexample; a square additionally needs right angles.
3. (iii) True. In parallelogram ABCD, compare △ABC and △BAD: AB common, BC=AD, and AC=BD. They are congruent by SSS, hence ∠ABC=∠BAD. These adjacent angles also sum to 180°, so each is 90°. It is a rectangle.

1. Given ∠DAC=∠CAB.
2. As AD ∥ BC, ∠DAC=∠ACB (alternate interior angles). Hence ∠CAB=∠ACB.
3. In △ABC, equal angles have equal opposite sides: AB=BC.
4. A parallelogram already has AB=CD and BC=AD, so AB=BC=CD=DA. Therefore ABCD is a rhombus.
5. As AB ∥ CD, ∠CAB=∠ACD; together with ∠ACB=∠CAB, we get ∠ACB=∠ACD. Hence AC also bisects ∠C.

1. (i) Yes. If each diagonal bisects the angles at both its endpoints, triangles on either side of a diagonal have matching base angles and a common base. This gives equal adjacent sides; all four become equal. (The statement means both diagonals bisect their respective endpoint angles.)
2. (ii) Yes. Mutual bisection first makes the quadrilateral a parallelogram. The perpendicular diagonals give equal adjacent sides by the Pythagorean theorem or SAS; it is a rhombus.
3. (iii) No. An isosceles trapezium can have equal diagonals without being a rectangle. Add that the diagonals bisect each other; then the shape is a parallelogram, and a parallelogram with equal diagonals is a rectangle.

1. Consecutive angles of a parallelogram sum to 180°. Their halves therefore sum to 90°.
2. When the bisectors of two consecutive angles meet, the angle between them is 90° (use the triangle formed with the original side).
3. The four intersections, taken in order, therefore form a quadrilateral with four right angles: a rectangle.
4. If AB = BC the parallelogram is a rhombus. Then the bisectors of ∠A and ∠C both lie along diagonal AC, and those of ∠B and ∠D along BD. All four bisectors pass through the centre, so the “four intersection points” are one single point and no rectangle is formed. That is why AB ≠ BC is assumed.

(i) By the Midpoint Theorem, PQ=BC/2, QR=AB/2 and RP=AC/2. Each of the four triangles △PQR, △APQ, △BPR and △CQR has side lengths AB/2, BC/2 and CA/2 (possibly in different orders). Therefore all four are congruent by SSS.

(ii) Yes, uniquely with the midpoint labels and orientation fixed. Draw through P a line parallel to QR, through Q a line parallel to PR, and through R a line parallel to PQ. The intersections of these three lines form the original triangle A,B,C: specifically A is the intersection through P and Q, B through P and R, C through Q and R. Each segment PQ, QR, RP is then a mid-segment.

1. By the Midpoint Theorem in △ABC, MN ∥ BC.
2. Since D lies on BC, MN is also parallel to BD.
3. Let MN meet AD at X. In △ABD, M is midpoint of AB and MX ∥ BD.
4. By the converse Midpoint Theorem, X is midpoint of AD: AX=XD.

1. Let E be midpoint of AD.
2. In △ADC, E and G are midpoints, so EG ∥ DC.
3. In △ADB, E and H are midpoints, so EH ∥ AB.
4. Since AB ∥ DC, the lines EG and EH both pass through E and are parallel to AB; they are therefore the same line. Thus G,E,H are collinear and GH ∥ AB.
5. If AB=CD as well as AB ∥ CD, then ABCD is a parallelogram. Its diagonals share one midpoint, so G=H and the “segment GH” has zero length; speaking of its direction would be meaningless.

(i) PQRS is a parallelogram (Varignon's theorem), and every parallelogram has diagonals that bisect each other. Therefore PR and QS bisect each other.

(ii) By the Midpoint Theorem, PQ=AC/2 and QR=BD/2. If AC=BD, then PQ=QR. A parallelogram with equal adjacent sides is a rhombus, and a rhombus has perpendicular diagonals. Hence PR ⟂ QS.

Converse: Yes. If PR ⟂ QS, then parallelogram PQRS is a rhombus, so PQ=QR. Thus AC/2=BD/2 and AC=BD.

(i) Construction with the original ABCD available: Make a congruent tracing of ABCD, then translate/rotate the tracing until its midpoints P,Q,R,S coincide with the copied parallelogram's P,Q,R,S in the corresponding order. Call the traced vertices A′,B′,C′,D′. Equivalently, choose a point A′ in the correct position and then successively reflect it in the midpoint points: B′=2P−A′, C′=2Q−B′, D′=2R−C′. The relation P+R=Q+S ensures S is the midpoint of D′A′.

(ii) A tracing or rigid motion preserves all distances and angles, so this particular copy is congruent to ABCD. Important qualification: PQRS alone does not determine ABCD uniquely up to congruence: the starting point A′ can be chosen anywhere. Arbitrary choices for A′ can give different shapes sharing the same four midpoints. To ensure congruence with the original, one must use some information about the original (for example its vertex placement or edge lengths).

(iii) The sides of PQRS satisfy PQ ∥ AC, QR ∥ BD, PQ=AC/2, QR=BD/2. Thus PQRS is a square iff adjacent sides PQ,QR are equal and perpendicular, which is exactly iff AC=BD and AC ⟂ BD. Both directions follow from these equivalences.

1. A regular pentagon has each interior angle equal to (5−2)×180°/5=108°.
2. At any ordinary vertex where pentagon corners meet, the total angle must be exactly 360°.
3. Three angles make 324°, leaving 36°. Four make 432°, causing overlap. No whole number of 108° corners makes 360°.
4. Allowing a vertex to touch the middle of another side does not fix the mismatch: a straight angle contributes 180°, and neither 180°+108° nor 180°+2×108° equals 360°.

Therefore a regular pentagon cannot tile the plane by congruent copies alone, although certain irregular pentagons can.

Draw any simple dart (one internal angle greater than 180°), trace many equal copies, and place one copy next to each exposed side by rotating 180° about that side's midpoint. Continue the pattern. The neighbours have exactly matching edges, and groups of four corners fill 360° around their meeting points. No gaps are left.

Alternatively, join the dart's side midpoints to obtain its Varignon parallelogram. Tile the plane with that parallelogram grid, align each dart with its midpoint parallelogram, and fill the spaces.

My choice: Method 1 is easiest to perform with paper cut-outs, because each step needs only an edge-midpoint half-turn. Method 2 gives a useful geometric explanation.

Take two congruent copies of △ABC. Rotate one copy through 180° about the midpoint of one of its sides. Together the two triangles form a parallelogram (matching opposite sides become equal and parallel). Translate copies of that parallelogram in rows and columns to cover the plane. Since each parallelogram consists of two copies of the original triangle, this gives a triangle tiling.

Count the equal gaps between the ruled lines from one end P of the segment to the other end Q. If there is an even number of gaps, the midpoint M is where the segment crosses the ruled line exactly halfway (for 4 gaps, the 2nd line from either end). If the number is odd, M lies halfway between the two middle lines.

Why it works: Drop the perpendicular from P to the ruled line through Q, meeting it at R. Because the lines are equally spaced, the middle ruled line passes through the midpoint of PR and is parallel to RQ. In △PRQ, by the converse of the Midpoint Theorem (Theorem 7), this line also passes through the midpoint of PQ.

Let θ=∠AED=∠BEC, the vertically opposite crossing angles. In △AED, ∠A+∠D=180°−θ. In △BEC, ∠B+∠C=180°−θ.

Therefore ∠A+∠B+∠C+∠D=360°−2θ < 360°, since θ>0.

Yes, 2° is possible. Make the lines AB and CD intersect so that θ=179°. Then the four end-angles add to 360°−358°=2°. The crossed figure is extremely narrow but valid.

Let O be the intersection of AC and BD. In parallelogram ABCD, O is the midpoint of BD, so DO=OB.

Rotate the figure through 180° about O. This swaps B and D and takes P to the unique point on BD whose distance from B is DP. Since BQ = DP, that point is Q. Therefore O is the midpoint of PQ, whichever of P and Q lies nearer B.

The diagonals of quadrilateral APCQ are AC and PQ. They bisect each other at O, so APCQ is a parallelogram for distinct P and Q. If both are O, the figure collapses to the segment AC.

1. When A is folded onto B, the crease is the perpendicular bisector of AB, so it meets AB at its midpoint P.
2. Since AB ⟂ BC, the crease through P is parallel to BC.
3. Let the crease meet hypotenuse AC at Q. By the converse Midpoint Theorem in △ABC, Q is the midpoint of AC.

So the single crease locates both required midpoints.

(i) Let P,Q,R be the midpoints of BC,CA,AB, and M the centroid. The medians cut the triangle into six equal-area little triangles. Pair △MPB with △MPC along the equal sides PB=PC, putting the two M vertices on opposite sides of the joined base. Because ∠MPB+∠MPC=180°, their outer edges become straight, forming a triangle with side lengths MB, MC and 2MP. Likewise pair the two pieces at Q and the two at R.

Since the centroid divides every median 2:1, each assembled triangle has the same three side lengths (2/3)AP, (2/3)BQ and (2/3)CR. They are therefore congruent by SSS.

(ii) Apply the same median-cut-and-pair construction to each of these three congruent triangles. Every new small triangle has sides two-thirds of the medians of the first-generation triangle. In the pictured assembly, P is the midpoint of MM′, so the segment from the joined vertex B = C to P is a median of length BC/2. The assemblies at Q and R show medians of lengths CA/2 and AB/2, opposite the sides of lengths 2MQ and 2MR respectively. By congruence, each assembled triangle has all three median lengths. Taking two-thirds of each gives AB/3, BC/3 and CA/3.

(iii) Following the hint: let P be the midpoint of BC and cut along the median AP. Then, in △APB, cut along the median PM, where M is the midpoint of AB. Turn △PMB through 180° about M: since M is the midpoint of AB, B lands on A, and P lands on a point P′ with M the midpoint of PP′. The two pieces △AMP and △AMP′ now form △APP′.

Compare △APP′ with △APC: AP is common; PP′ = 2PM = AC (by the Midpoint Theorem in △ABC, PM = AC/2); and AP′ = BP = PC. So △APP′ ≅ △APC by SSS: the triangle has been cut and reassembled into two congruent triangles.

(i) Let M be midpoint of diagonal BD. In △ADB, EM ∥ AB and EM=AB/2. In △DBC, the line through M parallel to DC meets BC at its midpoint F and MF=DC/2. Because AB ∥ DC, E,M,F lie on the same parallel line. Thus BF=FC and EF=EM+MF=(AB+CD)/2 (in the usual trapezium position shown).

(ii) Assume E and F are side midpoints. With M as midpoint of BD, the Midpoint Theorem gives EM ∥ AB in △DAB and FM ∥ DC in △DBC. Since AB ∥ DC, EM and FM are the same line; hence EF ∥ AB. The same length formula follows.

Since AB ∥ CD, in triangles △AOP and △COQ:

1. AO=OC (diagonals of a parallelogram bisect each other).
2. ∠AOP=∠COQ (vertically opposite angles).
3. ∠OAP=∠OCQ (alternate interior angles, AB ∥ CD).

Thus △AOP ≅ △COQ by ASA, and corresponding sides give OP=OQ. Since P,O,Q are collinear, O is the midpoint of PQ. An even shorter explanation uses 180° rotational symmetry about O, which maps AB onto CD and P onto Q.

The midline EF is parallel to both bases and has length (3+5)/2=4 cm. It lies halfway between the bases, so each smaller trapezium has the same height h/2, where h is the full height.

Top area = (3+4)/2 × h/2 = 7h/4.

Bottom area = (4+5)/2 × h/2 = 9h/4.

Therefore the area ratio is 7 : 9.

(i) Number of different cyclic orders of four labelled vertices, identifying a reversed tour as the same outline: (4−1)!/2=3.

(ii) There are two cases:

• All four points on the boundary of their convex hull: exactly 1 convex quadrilateral and 2 self-intersecting quadrilaterals.
• One point inside the triangle formed by the other three: 0 convex, 0 self-intersecting, and all 3 non-convex.

No third arrangement is possible if no three of the four points are collinear.

Parallel lines divide the two sides of a triangle in the same ratio. Thus AP/PB=AQ/QC, or QC=(PB/AP)×AQ. Since AP=1 and AQ=√5:

1. (i) PB=2: QC=2√5.
2. (ii) PB=1/3: QC=√5/3.
3. (iii) PB=2/3: QC=2√5/3.
4. (iv) Why for rational AP/PB? If AP/PB=m/n for positive integers m,n, divide AP into m equal pieces and PB into n equal pieces. Draw parallels to BC through the resulting division points. The corresponding equal parts on AC have the same length by repeated applications of the midpoint theorem/equal-intercept construction. Thus AQ contains m equal parts and QC contains n equal parts, making AQ/QC=m/n=AP/PB. Grade 10 generalises this to all real ratios (Basic Proportionality Theorem).

(i) Let the diagonals meet at O; O is the midpoint of both AC and BD. In △ABD, DM (M the midpoint of AB) and AO (O the midpoint of BD) are medians, so they meet at the centroid X of △ABD, with AX = ⅔AO = ⅓AC (Centroid Theorem). In the same way, in △CBD the medians BN and CO meet at Y with CY = ⅔CO = ⅓AC. So X and Y cut AC into three equal parts: DM and BN trisect AC.

(ii) Three constructions:

1. Parallelogram: build a parallelogram having PQ as a diagonal. Join two opposite corners to the midpoints of the opposite sides as in (i); the crossings divide PQ into thirds.
2. Equal-intercepts: draw an auxiliary ray at P, mark three equal steps on it, join the third mark to Q, then draw parallels through the first and second marks to meet PQ (the usual similar-triangles procedure, as in Ex. 11).
3. Centroid: make PQ a median of a triangle with P as vertex and Q as midpoint of the opposite side. Its centroid marks the point 2/3 from P. Obtain the other 1/3 point by doing the analogous construction with Q as vertex, or reflect the first trisection point in the midpoint of PQ.

(i) Yes, for concave and self-intersecting 4-gons. The Midpoint Theorem applies separately to △ABC, △BCD, △CDA and △DAB whether or not the original outline is crossed. It gives PQ ∥ RS ∥ AC and QR ∥ SP ∥ BD, so the midpoint 4-gon is a parallelogram unless degenerate.

(ii) Degeneration occurs precisely when AC ∥ BD: both pairs of sides of the midpoint figure lie on the same line, so the four midpoints are collinear (some may coincide). For an ordinary convex quadrilateral this cannot happen because its diagonals cross.

(iii) Yes, even for four non-coplanar points. Write P=(A+B)/2, Q=(B+C)/2, R=(C+D)/2, S=(D+A)/2 using position vectors. Then Q−P=(C−A)/2=R−S and R−Q=(D−B)/2=S−P. These vector equalities show P,Q,R,S lie in a plane and make a parallelogram, unless it degenerates.

No. For a counterexample, take a square A=(0,0), B=(4,0), C=(4,4), D=(0,4). Put P=(1,0), Q=(4,1), R=(3,4), S=(0,3): each is one quarter of the way along the respective side, not a midpoint.

We have PQ=(3,1), SR=(3,1), QR=(−1,3), PS=(−1,3). Both pairs of opposite sides are parallel and equal, so PQRS is a parallelogram. Thus the proposed converse of Varignon's theorem is false.

(i) Midpoint Theorem: Let P and Q be midpoints of AB and AC. Extend PQ through Q to S with QS=PQ. In triangles △APQ and △CSQ, AQ=CQ, PQ=QS, and ∠AQP=∠CQS (vertically opposite); hence the triangles are congruent by SAS. Thus AP=CS and AP ∥ CS. Since AP=PB, we have BP=CS and BP ∥ CS. Therefore BPSC is a parallelogram. Its opposite sides PS and BC are equal and parallel; as Q is midpoint of PS, PQ ∥ BC and PQ=BC/2.

(ii) Converse: Let P be midpoint of AB and let PQ ∥ BC meet AC at Q. Draw the line through C parallel to AB; let it meet PQ extended beyond Q at S. As BP ∥ CS and PS ∥ BC, BPSC is a parallelogram. Therefore CS=BP=AP. In △APQ and △CSQ, ∠AQP=∠CQS (vertically opposite), and ∠APQ=∠CSQ (AP ∥ CS). Together with AP=CS, the triangles are congruent by AAS. Hence AQ=QC, so Q is the midpoint of AC. The same congruence also gives PQ=QS, showing that the extension is by exactly one extra PQ length.

This is an open-ended investigation. The following examples show how to test converses:

General lesson: a property of a shape does not automatically characterise that shape unless its converse is proved.

Draw the diagonal AC. Triangles ABC and ADC are planar individually, although not in the same plane. The two triangle angle sums together are 360°.

At vertex A, the spatial angle between AB and AD is strictly less than ∠BAC+∠CAD if the two triangles meet with a genuine fold; the same holds at C. The angles at B and D are unchanged. Each angle at the fold is smaller than the matching sum in the flat triangles, so the four spatial internal angles add up to strictly less than 360°.

Yes, 2° can occur. Take two equal sticks AC and BD, centred one above the other and pointing in perpendicular horizontal directions. Join their ends in order A, B, C, D. All four joining edges have equal length, and all four corner angles are equal (each is opposite one of the equal sticks in an isosceles triangle). With the sticks nearly at the same level, each angle is close to 90°. As you raise one stick, the edges grow longer while the sticks keep their length, so each angle decreases continuously towards 0°. At some positive separation each angle is exactly ½°, giving a non-planar example with sum 2°.

Caution: Here “angle” means the smaller ordinary spatial angle between the two adjacent segments; a non-planar polygon does not enclose an ordinary planar interior region.

(i) Write the vertices of the original quadrilateral as S,O,M,E and let t=O−E be the vector of its diagonal EO. Translate every vertex by t: E′=E+t=O, O′=O+t, S′=S+t, M′=M+t. The four connector quadrilaterals formed by corresponding vertices and edges (e.g. E–S–S′–E′) are parallelograms, because each pair of connector edges is the same vector t. Translation preserves angles and lengths, so S′O′M′E′ is congruent to SOME and exactly matches the shifted cutout. This gives the required proof without relying on visual appearance.

(ii) Repeat such translations along the two diagonal vectors EO and SM, in positive and negative directions. The 9 coloured copies with displacement vectors −u,0,u combined with −v,0,v form a 3×3 lattice arrangement (here u=O−E and v=M−S). Four gaps between adjacent copies have boundaries matching four edges of SOME under suitable 180° rotations, so each gap can be filled by a congruent copy. Continue the lattice and the gap filling throughout the plane.

(i) A triangle with side lengths a≤b≤c exists exactly when a+b>c.

(ii) For a simple 4-gon, each side must be strictly shorter than the sum of the other three. With lengths 2,5,11,d:

• 11<2+5+d ⇒ d>4.
• d<2+5+11 ⇒ d<18.

Other inequalities are then automatically satisfied. Thus 4<d<18. Therefore 100: No; 10: Yes; 1: No.

(iii) For any four positive a,b,c,d, a non-degenerate quadrilateral exists exactly when max(a,b,c,d) < (a+b+c+d)−max(a,b,c,d). This is the quadrilateral inequality. It is necessary because one side cannot reach farther than the other three joined end-to-end; it is sufficient because the joint between the other three can be adjusted continuously to meet the largest side.

A diagonal joins two non-adjacent vertices of a polygon.

From each vertex, there are n−3 possible diagonal endpoints (exclude itself and its 2 neighbours). This counts every diagonal twice, once from each end. Therefore:

When n increases to n+1, the number added is D(n+1)−D(n)=n−1.

Any simple n-gon can be divided into n−2 triangles by drawing non-crossing diagonals between suitable vertices. Each triangle has angle sum 180°. Adding the triangle angles counts the polygon's internal angles exactly once overall.

Checks: n=3 gives 180°, n=4 gives 360°, n=5 gives 540°, n=6 gives 720°. The proof works for concave simple polygons too, using triangulation rather than necessarily drawing every diagonal from a single vertex.

Assume P is on AB and Q on AC.

1. (i) PRLL + HALF ⇒ P MID + Q MID: True. Through Q draw the line parallel to AB, meeting BC at R. Then PQRB has both pairs of opposite sides parallel, so it is a parallelogram and BR = PQ = BC/2: R is the midpoint of BC. In △ABC the line through the midpoint R parallel to AB meets AC at its midpoint (Theorem 7), so Q is the midpoint of AC. Finally, the line through Q parallel to BC meets AB at its midpoint (Theorem 7 again), so P is the midpoint of AB.
2. (ii) P MID + PRLL ⇒ Q MID + HALF: True. This is the Converse Midpoint Theorem. Once Q is midpoint, the usual Midpoint Theorem also gives PQ=BC/2.
3. (iii) P MID + HALF ⇒ Q MID or PRLL: Not always. A segment of the correct length need not be parallel to BC. For a concrete counterexample, take A=(0,0), B=(4,0), C=(4,2), P=(2,0), and Q=(1.2,0.6) on AC. Here BC=2 and PQ=√[(−0.8)²+0.6²]=1=BC/2, while AP=PB=2. But AQ/AC=0.3, not 0.5, and PQ is not parallel to BC. So neither conclusion follows.

The important idea is that two pieces of information in a theorem are not always interchangeable when forming a converse.

Choose a diagonal AC that lies inside the tile and draw it horizontally, with B above it and D below. This works for a concave tile too: use the diagonal from its inward corner. AC cuts the tile into two triangles.

1. Build a row. Translate the tile repeatedly by AC in both directions. The horizontal bases AC now meet end to end. Above them are copies of △ABC, all pointing up. Between consecutive tips B, draw a horizontal segment. Each gap below this segment is a downward triangle congruent to △ABC: a half-turn about the midpoint of the shared sloping side puts it exactly in place. These upright and inverted triangles fill the whole strip, edge to edge. The same argument fills the strip below AC with copies of △ADC.
2. Stack the rows. Shift the starting row by the vector from D to B. The D tips of the new row land exactly on the B tips of the old row. Along each segment joining consecutive tips, the inverted △ABC from below and the inverted △ADC from above share a side equal to AC. Together they form one half-turned copy of ABCD. Repeat the shift up and down. Every strip is filled once, so there are no gaps or overlapping interiors.
3. Relate this to the three methods. The unturned tiles are the copies reached by the two diagonal translations AC and DB (Method 3). The tile across each edge is its half-turn about that edge’s midpoint (Method 1). For Method 2, the two sides of the midpoint parallelogram are parallel to AC and DB and half their lengths. Moving two grid steps in either direction therefore places an unturned tile in precisely the position given by Method 3. The spaces between them are the half-turned tiles just constructed.

Answer: 1/5 of the square.

Each of the four lines joins a corner to the midpoint of a side, and together they cut out a central square (shaded). Look at one midpoint: a small right triangle sits between the line ending there and the side of the big square. Turn that small triangle through 180° about the midpoint. It lands outside the big square and, together with the piece next to the shaded square, completes a square exactly equal to the shaded one (see the figure).

Doing this at all four midpoints rearranges the whole big square, with nothing added or lost, into 5 equal squares: the shaded one and four around it. So the shaded square is 1/5 of the big square, whatever its size.