Quadrilaterals

AD and BC share endpoints with AB.

CD shares no endpoint with AB.

∠B and ∠D share sides with ∠A.

∠C is across the quadrilateral.

Opposite sides are two sides with no common endpoint.

Opposite angles are angles whose vertices are not joined by a side.

At vertex A, the two adjacent sides AB and AD form two angles.

Choose the angle whose interior contains the opposite vertex C.

A non-convex indentation would place the two connecting sides so the diagonals do not cross while the ordered opposite sides cannot remain parallel without overlap or intersection.

Equality of one opposite-side pair gives no direction information.

A dart-shaped non-convex quadrilateral can be drawn with those two sides equal.

A dart can be arranged symmetrically so its opposite angles A and C are equal while the angle at the dent exceeds 180°.

First trace AB, BC, CD, DA and test whether non-adjacent sides cross.

If they cross, it is self-intersecting; otherwise test whether D lies inside triangle ABC.

Inside gives a non-convex quadrilateral; in the exterior region opposite B across AC gives the convex order ABCD, while the other exterior regions are classified by the crossing test.

Self-intersection means two non-adjacent segments share an interior point.

Two intersecting lines determine a plane; their four endpoints lie in that plane.

Adjacent angles are supplementary and opposite angles are equal.

Starting with 90°, all four angles become 90°.

Every rhombus already has perpendicular diagonals.

A non-square rhombus is therefore a counterexample.

Let diagonals AC=BD. Compare triangles ABC and BAD: AB common, BC=AD, AC=BD.

SSS congruence gives ∠ABC=∠BAD; adjacent parallelogram angles are also supplementary, so each is 90°.

AB∥CD gives ∠BAC=∠DCA; AD∥BC gives ∠DAC=∠BCA.

Given ∠BAC=∠DAC, hence ∠DCA=∠BCA, so AC bisects ∠C.

Triangles BAC and DAC are congruent by ASA, giving AB=AD; with opposite sides equal, all four sides match.

No for a general quadrilateral: a kite has one diagonal bisecting two angles but not necessarily both diagonals.

Add that each diagonal bisects both endpoint angles; then adjacent triangles yield all four sides equal.

Bisecting each other makes the quadrilateral a parallelogram.

Perpendicular diagonals in that parallelogram make adjacent sides equal.

An isosceles trapezium has equal diagonals but need not be a rectangle.

Add the condition that the diagonals bisect each other, making a parallelogram; then equal diagonals imply rectangle.

Adjacent angles of a parallelogram sum to 180°, so their half-angles sum to 90°; adjacent bisectors are perpendicular.

Opposite angles are equal, so their bisectors are parallel.

Thus the intersection quadrilateral has four right angles and opposite sides parallel.

PQ=BC/2=BR=RC; PR=AC/2=AQ=QC; QR=AB/2=AP=PB.

Each of ΔPQR, ΔQPA, ΔRPB and ΔQRC has side lengths AB/2, BC/2, AC/2.

Reflect R across P to get B, P across Q to get A, and Q across R to get C; equivalently construct the three outer congruent triangles.

Join A,B,C.

MN∥BC, hence MN∥BD in triangle ABD.

M is midpoint of AB, so by the converse midpoint theorem MN meets AD at its midpoint.

Let E be midpoint of AD. In triangle ADC, EG∥DC; in triangle ADB, EH∥AB.

Since AB∥DC, EG and EH are the same line through E.

PQRS is a parallelogram by Varignon’s theorem.

Diagonals of a parallelogram bisect each other.

In Varignon parallelogram, PQ=AC/2 and QR=BD/2.

AC=BD makes adjacent sides PQ=QR, so PQRS is a rhombus; its diagonals PR,QS are perpendicular.

Conversely, perpendicular diagonals make parallelogram PQRS a rhombus, so PQ=QR and AC=BD.

Choose A′ so S and P are midpoints of A′D′ and A′B′; reflect A′ across P and S to obtain B′ and D′.

Reflect B′ across Q to get C′ and verify its reflection across R is D′.

The same midpoint reflections preserve the vectors between consecutive vertices.

Triangles around each midpoint, such as ΔSDR and ΔSD′R, match by SAS; corresponding sides and angles follow around the boundary.

PQ∥AC with PQ=AC/2; QR∥BD with QR=BD/2.

Square gives PQ=QR and PQ⊥QR, hence AC=BD and AC⊥BD.

Conversely equal perpendicular diagonals make adjacent Varignon sides equal and perpendicular, so PQRS is a square.

Each interior angle is 108°.

At a tiling vertex the angles must total 360°, but 360/108=10/3 is not an integer.

Place a copy beside the dart and rotate it 180° about the midpoint of the shared edge.

Repeat across every exposed edge; four angles meet around repeated vertices and total 360°.

Join two congruent copies along one side after a half-turn; their union is a parallelogram.

Tile the plane with that parallelogram grid.

Place the segment across two convenient guide lines. Through each endpoint draw a transversal to a middle guide so the construction forms a trapezium.

The line halfway between the endpoint guide levels meets the segment at its midpoint by the trapezium midpoint theorem.

At crossing E, triangles ABE and CDE give A+B=180°−∠AEB and C+D=180°−∠CED.

The two crossing angles are equal and positive, so total=360°−2∠AEB<360°.

Choose the crossing angle near 179° to make the total near 2°.

Diagonals AC and BD meet at O and bisect each other.

Since DP=BQ symmetrically on BD, O also bisects PQ.

Thus diagonals AC and PQ of APCQ bisect each other.

The crease is the perpendicular bisector of AB, so it crosses AB at its midpoint M and is perpendicular to AB.

The right angle is at B, so BC is also perpendicular to AB; therefore the crease through M is parallel to BC.

By the converse midpoint theorem, its intersection N with AC is the midpoint of AC.

Rotate ΔMPB by 180° about P and join it to ΔMPC along PB=PC. The rotated M is M′, and M,P,M′ are collinear, so the union is triangle MCM′.

Its sides are MM′=2MP=AM, MC=CM and CM′=BM, using the centroid’s 2:1 division.

Repeat about Q and R. Each new triangle has side lengths AM, BM and CM.

Apply the same median dissection to each congruent triangle.

All scale by another factor of 1/3 in the relevant reassembly.

Draw median AP of ΔABC. In ΔAPB, let R be midpoint of AB and cut along median PR.

Rotate ΔRAP by 180° about R. It joins ΔRBP into a triangle whose sides are AP, BP and 2PR.

Since BP=PC and PR=AC/2 by the midpoint theorem, these sides are AP, PC and AC, matching ΔAPC.

Draw diagonal BD and let M be its midpoint. In ΔDAB, EM∥AB and EM=AB/2.

In ΔDBC, the line through M parallel DC meets BC at its midpoint F and MF=DC/2.

E,M,F are collinear, so EF=EM+MF=(AB+CD)/2.

With M midpoint of diagonal BD, EM∥AB in ΔDAB and FM∥DC in ΔDBC.

Since AB∥DC, both are the unique line through M in that direction.

In parallelogram ABCD, OA=OC and AB∥CD.

Triangles OAP and OCQ are congruent by ASA using vertical and alternate angles.

Therefore OP=OQ.

Mid-segment EF=(3+5)/2=4 cm and lies halfway between bases.

Both smaller trapeziums have equal height h/2.

Areas are (3+4)h/4 and (4+5)h/4.

There are 4! orders; divide by 4 rotations and 2 directions.

4!/(4×2)=3.

If all four points are on their convex hull: one order is convex and two are self-intersecting.

If one point lies inside the triangle of the other three: all three simple orders are non-convex, with none convex or self-intersecting.

Similarity gives AP/PB=AQ/QC.

Substitution gives QC=10.

Similarity gives AP/PB=AQ/QC.

Substitution gives QC=5/3.

Similarity gives AP/PB=AQ/QC.

Substitution gives QC=10/3.

Write AP/PB=m/n. Divide AP into m equal parts and PB into n equal parts.

Draw parallels through the division points; the equal-intercept theorem gives matching equal divisions on AC.

AQ contains m parts and QC contains n parts.

Use vectors A=0, B=b, D=d, C=b+d; then M=b/2 and N=d+b/2.

Line DM meets AC at parameter 1/3, while BN meets it at 2/3.

Make PQ a parallelogram diagonal and use the midpoint construction from part (i).

Alternatively use the rational parallel-line construction with a ray divided into three equal units.

A third method constructs a triangle having PQ as a median and uses its centroid, then repeats from the other end.

In triangles ABC and ADC, midpoint segments PQ and SR are both parallel to AC.

In triangles BCD and BAD, QR and PS are both parallel to BD.

Its adjacent sides are parallel to diagonals AC and BD.

It has zero area exactly when those diagonal directions are parallel.

Each triple of vertices still forms a plane triangle, so its midpoint segment theorem applies.

The opposite midpoint segments are parallel in 3D, which also forces P,Q,R,S into one plane.

Take unit square A(0,0),B(1,0),C(1,1),D(0,1).

For any 0<t<1, choose P(t,0), Q(1,t), R(1−t,1), S(0,1−t). The opposite side vectors match.

For t≠1/2 these points are not midpoints.

Extend PQ to S so QS=PQ.

AQ=CQ, vertical ∠AQP=∠CQS, and PQ=QS, so ΔAPQ≅ΔCSQ by SAS.

Thus AP=CS and AP∥CS; since AP=PB, PB=CS and PBCS is a parallelogram.

Extend PQ to S with QS=PQ. Given AP=PB and PQ∥BC.

The construction gives a parallelogram PBCS, so CS=PB=AP and CS∥AP.

Then quadrilateral AP?C forces AQ=QC by the same SAS/parallel argument.

True examples: parallelogram with perpendicular diagonals → rhombus; parallelogram with equal diagonals → rectangle; parallelogram with equal perpendicular diagonals → square.

False without “parallelogram”: equal diagonals may give an isosceles trapezium, and perpendicular diagonals may give a kite.

Use AC as a hinge joining triangles ABC and ADC. When coplanar on opposite sides, their corner angles combine to 360°.

Folding about AC makes the spatial angles at A and C strictly smaller by the triangle inequality for angles; B and D stay fixed.

Start with very thin triangles and fold almost flat in the same direction to make all four corner angles total as close to 0° as desired, including 2°.

The four midpoint/diagonal pieces around each copy form parallelograms by Varignon’s theorem.

Opposite sides in those parallelograms are equal and parallel, so translation along EO maps E,O,M,S to E′,O′,M′,S′.

Translate by each of the two diagonal vectors in positive and negative directions.

Translations commute, so reaching a grid position in either order gives the same copy.

Neighbouring copies meet at vertices; the intervening boundary cycles repeat the original side vectors.

Order a≤b≤c.

The necessary and sufficient condition is a+b>c.

Largest-side condition requires every side be less than the sum of the other three.

For 11: 11<2+5+d gives d>4; for d: d<2+5+11=18.

Find M=max(a,b,c,d) and total T=a+b+c+d.

Check M<T−M.

A diagonal joins two non-adjacent vertices.

Counts for n=3,4,5,6,7,8 are 0,2,5,9,14,20.

There are n(n−3) ordered vertex-to-non-neighbour connections.

Divide by 2 because each diagonal has two endpoints.

Adding one vertex increases the count by n−1.

Draw diagonals from one vertex to split a convex n-gon into n−2 triangles.

For a non-convex simple polygon, an ear decomposition gives the same count.

Triangles APQ and ABC are similar.

Their scale factor is PQ/BC=1/2, so AP/AB=AQ/AC=1/2.

Keep P MID fixed and reverse “Q MID implies PRLL”.

The converse is: P MID and PRLL imply Q MID; HALF also follows.

No. Let A=(0,0), B=(2,0), C=(3,1), so P=(1,0).

BC/2=√2/2. Besides midpoint (1.5,0.5), the point Q=(0.3,0.1) on AC also satisfies PQ=√2/2.

This Q is not the midpoint and PQ is not parallel to BC.

Half-turn method: paired copies share full edges, and the four vertex angles sum to 360°.

Varignon method: the midpoint parallelograms tile as a lattice; congruent quadrilaterals fill the cells and gaps edge-for-edge.

Translation method: two diagonal translations generate a lattice; commuting translations align every shared boundary.

Use a unit square. The four lines are 2x+y=1, 2x+y=2, −x+2y=0 and −x+2y=1.

Their central intersections are (1/5,3/5), (3/5,4/5), (4/5,2/5), (2/5,1/5).

The shoelace formula gives area 1/5.