A quadrilateral must be a closed figure made of exactly four straight line segments that meet at four vertices, with no two sides crossing each other.

Figures (i), (ii), (iii) satisfy this — four straight sides, closed, non-intersecting.

Figure (iv) is not a quadrilateral because one of its sides is curved, not straight.

Figure (v) is not a quadrilateral because it is a self-intersecting (crossed) figure — two of its sides cross each other, so it isn’t a simple closed figure.

Compare $\triangle ADC$ and $\triangle DAB$. Since ABCD is a rectangle:

• $AB = CD$ (opposite sides of rectangle)
• $\angle BAD = \angle CDA = 90^\circ$
• $AD$ is common to both triangles

By the SAS congruence condition: $\triangle ADC \cong \triangle DAB$

Since $AC$ and $BD$ are corresponding parts of these congruent triangles, $AC = BD$.

The blue angles $\angle 1$ (at O between OA, OB) and the angle between OC, OD are equal — they are vertically opposite angles.

Label $\angle 1 = \angle OBA$ and $\angle 2 = \angle ODC$, and let $\angle 3 = \angle OBC$.

Since $\angle B = 90^\circ$:   $\angle 3 + \angle 1 = 90^\circ$

In $\triangle BCD$:   $\angle 3 + \angle 2 + 90^\circ = 180^\circ \Rightarrow \angle 3 + \angle 2 = 90^\circ$

Comparing the two: $\angle 1 = 90^\circ – \angle 3 = \angle 2$

By the AAS condition: $\triangle AOB \cong \triangle COD$

So $OA = OC$ and $OB = OD$ (corresponding parts of congruent triangles).

The four angles around O formed by two crossing lines come in vertically-opposite pairs: $60^\circ, 60^\circ, 120^\circ, 120^\circ$ (since $60+120=180^\circ$, a linear pair).

In $\triangle AOB$, since the diagonals bisect each other, $OA = OB$. So it’s isosceles, and its base angles are equal — call each one $a$:

Similarly every triangle formed ($\triangle BOC$, $\triangle COD$, $\triangle DOA$) is isosceles with apex angle $120^\circ$ or $60^\circ$, giving base angles of $30^\circ$ or $60^\circ$ as worked out in the figure. Each vertex angle of the quadrilateral becomes $30^\circ + 60^\circ = 90^\circ$.

Since $\triangle AOB \cong \triangle COD$ and $\triangle AOD \cong \triangle COB$, we get $AB = CD$ and $AD = CB$.

The four angles at O are $x, x, 180^\circ – x, 180^\circ – x$ (vertical pairs + linear pair).

In isosceles $\triangle AOB$ (base angles $a$):   $2a + x = 180^\circ \Rightarrow a = 90^\circ – \dfrac{x}{2}$

In isosceles $\triangle AOD$ (base angles $b$):   $2b + (180^\circ-x) = 180^\circ \Rightarrow b = \dfrac{x}{2}$

Each vertex angle of the quadrilateral $= a + b$:

Join diagonal $BD$. Compare $\triangle BAD$ and $\triangle DCB$. Let $\angle 1 = \angle ABD$, $\angle 2 = \angle BDC$, $\angle 3 = \angle DBC$.

Since $\angle B = 90^\circ$:   $\angle 3 + \angle 1 = 90^\circ$

In $\triangle BCD$, since $\angle C = 90^\circ$:   $\angle 3 + \angle 2 + 90^\circ = 180^\circ \Rightarrow \angle 3+\angle2=90^\circ$

So $\angle 1 = \angle 2$.

We also have $BD$ common to both triangles, and $\angle A = \angle C = 90^\circ$. By the AAS condition: $\triangle BAD \cong \triangle DCB$

Consider $AD$ as the transversal cutting $AB$ and $DC$.

$\angle A + \angle D = 90^\circ + 90^\circ = 180^\circ$ — these are interior angles on the same side of transversal $AD$.

We already know: equal diagonals + bisecting each other → rectangle (90° angles, opposite sides equal). To additionally get all four sides equal, we need the right angle between the diagonals.

In square ABCD with diagonals meeting at O, compare $\triangle BOA$ and $\triangle BOC$:

• $BA = BC$ (all sides of a square are equal)
• $OA = OC$ (diagonals bisect each other)
• $BO$ is common

By the SSS condition: $\triangle BOA \cong \triangle BOC$, so $\angle BOA = \angle BOC$ (corresponding parts).

Since $\angle BOA$ and $\angle BOC$ together form a straight line:   $\angle BOA + \angle BOC = 180^\circ$

Since the two are equal:   $2\angle BOA = 180^\circ \Rightarrow \angle BOA = 90^\circ$

In $\triangle ADC$:   $\angle 1 + \angle 3 + 90^\circ = 180^\circ$. Since $AD = DC$ (square), the angles opposite them are equal: $\angle 1 = \angle 3$.

By identical reasoning in $\triangle ABC$ (with $AB = BC$): $\angle 2 = \angle 4 = 45^\circ$.

Draw diagonal $SM$. This splits quadrilateral SOME into $\triangle SEM$ and $\triangle SOM$.

Angle sum of a triangle is $180^\circ$, so:

Adding both:

Regrouping, $(\angle1+\angle4) + (\angle3+\angle6) + \angle2 + \angle5 = 360^\circ$, and these four bracketed groups are exactly the four vertex angles of quadrilateral SOME.

Deduction 6 (Angles): Since $AB \parallel CD$ and $AD$ is a transversal:

Since $AD \parallel BC$, with $AB, CD$ as transversals: $\angle A + \angle B = 180^\circ$ and $\angle C + \angle D = 180^\circ$, giving:

Deduction 7 (Sides): In $\triangle ABD$ and $\triangle CDB$: the angles marked with a single arc are equal (opposite angles of the parallelogram), and the angles marked with double arcs are equal (alternate angles, since $AD \parallel BC$ with transversal $BD$). By AAS: $\triangle ABD \cong \triangle CDB$.

Let $\angle P = x$ in parallelogram PERA (vertices P, R, A, E around). Since $\angle P + \angle R = 180^\circ$ (co-interior angles):

Since $\angle A + \angle R = 180^\circ$ as well:

Compare $\triangle AOE$ and $\triangle YOS$ (diagonals AY and ES meet at O):

• $AE = YS$ (opposite sides of the parallelogram)
• Single-arc angles equal — alternate angles ($AE \parallel SY$, transversal AY)
• Double-arc angles equal — alternate angles ($AE \parallel SY$, transversal ES)

By the ASA condition: $\triangle AOE \cong \triangle YOS$

Draw diagonal $AE$, splitting GAME into $\triangle GAE$ and $\triangle MAE$. Label the four angles at A and E as $a, b, c, d$.

In $\triangle GAE$: since $GE = GA$ (rhombus sides), the angles opposite them are equal: $a = d$.

In $\triangle MAE$: since $ME = MA$, similarly $b = c$.

$\triangle GAE \cong \triangle MAE$ by SSS (all 4 rhombus sides equal, $AE$ common), so $a = b$, $c = d$, and $\angle G = \angle M$.

Applying this to rhombus ABCD (one angle 50°): the diagonal creates 4 equal angles $a$ at the two 50°-corners’ triangle. In $\triangle ADB$: $a + a + 50^\circ = 180^\circ \Rightarrow a = 65^\circ$.

In rhombus GAME with diagonals meeting at O, compare $\triangle GEO$ and $\triangle MEO$:

• $GE = ME$ (sides of rhombus)
• $GO = MO$ (diagonals of a parallelogram bisect each other — Deduction 8 applies, since rhombus is a parallelogram)
• $EO$ is common

By SSS: $\triangle GEO \cong \triangle MEO$

$\angle GOE = \angle MOE$ (corresponding parts), and they form a linear pair: $\angle GOE + \angle MOE = 180^\circ \Rightarrow$ each $= 90^\circ$.

Compare $\triangle ABD$ and $\triangle CBD$:

• $AB = CB$ (given)
• $AD = CD$ (given)
• $BD$ is common

By SSS: $\triangle ABD \cong \triangle CBD$

So $\angle ABD = \angle CBD$ and $\angle ADB = \angle CDB$ (corresponding parts) — i.e. BD bisects $\angle ABC$ and $\angle ADC$.

Now let BD meet AC at O. Compare $\triangle AOB$ and $\triangle COB$:

• $AB = CB$ (given)
• $\angle ABO = \angle CBO$ (just shown, BD bisects $\angle B$)
• $BO$ is common

By SAS: $\triangle AOB \cong \triangle COB$

So $AO = CO$ (BD bisects AC), and $\angle AOB = \angle COB$. Since these form a linear pair summing to $180^\circ$, each is $90^\circ$.

Since $XW \parallel UV$ and $XY \perp UV$: $\angle XYZ = 90^\circ$, so $a = 180^\circ – 90^\circ = 90^\circ$. Similarly $b = 90^\circ$. Also $XY \parallel WZ$ (both perpendicular to UV) and $XW \parallel YZ$ (both lie along/parallel to UV-direction).

Compare $\triangle UXY$ and $\triangle VWZ$:

• $UX = VW$ (given, isosceles trapezium)
• $XY = WZ$ (opposite sides of rectangle XWZY)
• $\angle XYU = \angle WZV = 90^\circ$

By RHS (right angle–hypotenuse–side): $\triangle UXY \cong \triangle VWZ$

Diagonals of a rectangle are equal and bisect each other, so each half-diagonal triangle is isosceles. Given $\angle ABD = 30^\circ$:

Reasoning: $\angle DAB = \angle ABC = 90^\circ$ (rectangle angles). Since $AB\parallel DC$, alternate angles on transversal $BD$ give $\angle ABD = \angle BDC = 30^\circ$. In right $\triangle ABD$: $\angle ADB = 90^\circ – 30^\circ = 60^\circ$. Similarly, using diagonal $AC$ (alternate angles on transversal $AC$): $\angle CAD = \angle ACB = 60^\circ$, and then $\angle ACD = 90^\circ – 60^\circ = 30^\circ$.

Given the angle at O (say $\angle POS$) $= 110^\circ$. Since diagonals bisect each other, all four triangles at O are isosceles.

Draw line segment $AB = 8$ cm. Mark its midpoint $O$.

At $O$, draw a line $l$ making the required angle (30°/40°/90°/140°) with $AB$.

On line $l$, mark points $C$ and $D$ on either side of $O$ such that $OC = OD = 4$ cm (half of 8 cm).

Join $AC, CB, BD, DA$. The resulting quadrilateral ACBD has equal diagonals bisecting each other at the chosen angle.

$PL$ and $AM$ are diameters, so they pass through centre $O$ and $OP=OL=OA=OM=$ radius $r$. So both diagonals of APML are equal (each $=2r$) and bisect each other at $O$.

They’re also given to be perpendicular ($90°$).

Let $AB$ and $CD$ be the two equal sticks. Place them so their midpoints coincide at a common point $O$ (like a letter X / a plus sign).

Tie the thread from $A$, pass it through $C$, then to $B$, then to $D$, and back to $A$ — forming quadrilateral $ACBD$.

Since the sticks are equal ($AB = CD$) and their midpoints coincide ($O$ bisects both), the diagonals of $ACBD$ are equal and bisect each other.

In parallelogram PRAE, adjacent angles are supplementary, opposite angles equal.

In parallelogram PQRS, with $\angle P = 110^\circ$:

In rhombus UVWX, diagonal $XV$ bisects the vertex angles (Deduction 9), so $\angle UXV = \angle WXV = 30^\circ$ too (by the symmetry of the rhombus’s diagonal property), making $\angle XVU = \angle XVW = \angle UXV = \angle WXV = 30^\circ$.

In $\triangle UVX$: $\angle U = 180^\circ – 30^\circ – 30^\circ = 120^\circ$. By rhombus symmetry, $\angle W = \angle U = 120^\circ$.

Given $\angle OEI = 20^\circ$, and by the kite-style symmetry shown in the figure, $\angle AOE = \angle EOI = 20^\circ$.

This gives $\angle A = \angle I = 140^\circ$ following the same angle-sum reasoning as the other isosceles triangles in the figure.

Draw line segment $AB = 7$ cm. Mark its midpoint $O$.

At $O$, draw a line $l$ making a $140^\circ$ angle with $AB$.

On line $l$, mark $C$ and $D$ on either side of $O$ such that $OC = OD = 2.5$ cm (half of 5 cm).

Join $AC, CB, BD, DA$.

Draw line segment $AB = 5$ cm. Mark midpoint $O$.

At $O$, draw a line perpendicular to $AB$ (since rhombus diagonals meet at exactly 90° — Deduction 10).

On the perpendicular, mark $C$ and $D$ on either side of $O$ with $OC = OD = 2$ cm (half of 4 cm).

Join $AC, CB, BD, DA$.

Every angle of an equilateral triangle is $60^\circ$. Joining two such triangles along a common side (length 4 cm) gives a quadrilateral with all 4 outer sides $= 4$ cm.

At each of the two “side” vertices, two $60^\circ$ angles combine: $60^\circ+60^\circ=120^\circ$. At the top and bottom vertices, the angle stays $60^\circ$ (one full triangle angle each).

Recall: in a kite, one diagonal perpendicularly bisects the other (Kite Property 1), but is itself not necessarily bisected.

Draw line segment $PQ = 6$ cm (the diagonal that gets bisected). Construct its perpendicular bisector — let it meet $PQ$ at $T$.

On this perpendicular line, mark points $R$ and $S$ on either side of $T$ (at any convenient unequal distances from $T$) such that $RS = 8$ cm total.

Join $PR, RQ, QS, SP$.

Since $PQ \parallel SR$, co-interior angles formed with each transversal add up to $180^\circ$. The two marked base angles at S and R are $135^\circ$ and $105^\circ$.

Since $AD = BC$, $ABCD$ is an isosceles trapezium, so angles at the base (opposite the equal sides) are equal (proved earlier): $\angle B = \angle A$. Given the angle at D is $100^\circ$ adjacent to side AD:

Rhombus and Square. A rhombus is the shape where the kite’s two pairs of equal adjacent sides become all four sides equal — at that point it gains parallel opposite sides too, becoming a parallelogram. A square is the special case that’s also a rectangle.

No — except for the square. A rectangle’s adjacent sides are generally unequal, which contradicts a kite’s requirement of two pairs of equal adjacent sides, unless all 4 sides are equal (i.e., it’s a square).

No, not every kite is a rhombus. A kite only needs two pairs of adjacent equal sides (e.g. 6,6,9,9) — these don’t have to all be equal. A rhombus is a special, more restrictive kite where all four sides happen to be equal. So every rhombus is a kite, but a kite need not be a rhombus.

PAIR and RODS are both rectangles sharing the corner $R$. Diagonal $RA$ of rectangle PAIR passes through point $O$, which is also a vertex of rectangle RODS. The angle between $RP$ and the diagonal $RA$ is given as $30^\circ$, i.e. $\angle PRA = 30^\circ$.

Since $PAIR$ is a rectangle, $PA \parallel RI$. Treating $RA$ as a transversal: $\angle PRA = \angle RAI$ (alternate angles), but more directly, since $RODS$’s side $RO$ lies along the same line as $RI$ (both horizontal in the figure), and $OD$ is parallel to $RI$ as well (opposite sides of rectangle RODS):

Draw line segment $AB = 6$ cm — this will be one diagonal.

Using a compass (no protractor needed), construct the perpendicular bisector of $AB$. Let it cross $AB$ at $O$.

On this perpendicular, mark points $C$ and $D$ on either side of $O$ such that $OC = OD = 3$ cm (half of 6 cm).

Join $AC, CB, BD, DA$.

Let the side of square CASE be $x$, so each half-side is $\dfrac{x}{2}$.

In right $\triangle CUV$ (at corner C), $CU = CV = \dfrac{x}{2}$ and $\angle C = 90^\circ$. By Pythagoras:

By identical reasoning at each corner: $UV=VW=WX=XU=\dfrac{x}{\sqrt2}$ — all four sides of UVWX are equal.

Since $\triangle CUV$ is right-angled isosceles, $\angle CUV = \angle CVU = 45^\circ$. The angle of UVWX at U is then:

This holds at every vertex of UVWX.

A quadrilateral with four equal sides is, by definition, a rhombus. Every rhombus is also a parallelogram (Deduction 9, note), so its adjacent angles are supplementary and opposite angles are equal.

If one angle is $90^\circ$, the angle adjacent to it must be $180^\circ – 90^\circ = 90^\circ$, and the angle opposite the first $90^\circ$ angle also equals $90^\circ$. This forces all four angles to be $90^\circ$.

Let ABCD have $AB=CD$ and $BC=AD$. Draw diagonal $AC$, splitting it into $\triangle ABC$ and $\triangle ACD$ (renamed $\triangle CDA$).

In $\triangle ABC$ and $\triangle CDA$: $AB=CD$, $BC=DA$, and $AC$ is common. By SSS: $\triangle ABC \cong \triangle CDA$.

So $\angle 1 = \angle 4$ and $\angle 2 = \angle 3$ (corresponding parts). Since $\angle 1, \angle 4$ are alternate angles for lines $AB, CD$ cut by transversal $AC$: $AB \parallel CD$. Since $\angle 2, \angle 3$ are alternate angles for lines $BC, AD$ cut by transversal $AC$: $BC \parallel AD$.

Join $BD$. This splits the quadrilateral into $\triangle ABD$ and $\triangle BDC$ — exactly the same strategy as for a convex quadrilateral.

Angle sum of a triangle is always $180^\circ$, regardless of shape:

The reflex angle at $D$ (the “dent”) is exactly the sum of the two angles each triangle contributes at $D$ — so the same regrouping argument as for SOME applies.