Chapter 1: Geometric Twins (Congruence)

No. With only the two lengths, the arms can open at any angle, giving many different symbols. Fixing the angle between them as well fixes the shape and size.

Drawing: draw BC = 8 cm; at B make an angle of 80° with a protractor; mark A on the new arm with BA = 4 cm.

No, the angle between the arms is also needed; with AB = 4 cm, BC = 8 cm and ∠B = 80° the symbol is fixed.

Measuring them, the two arms have the same lengths in both figures (about 1.7 cm and 3.4 cm), but the angles between the arms are different: about 90° in the first and about 70° in the second. So the figures are not congruent: one cannot be placed exactly over the other.

Not congruent: equal arm lengths but different angles (about 90° and 70°).

• Drops: congruent (the second is the first turned on its side).
• Clouds: not congruent (the second cloud is smaller).
• Stars: not congruent (different sizes).
• Leaves: congruent (one pair is the mirror image of the other; flipping is allowed).

The drops and the leaves are congruent pairs; the clouds and the stars are not.

(a) A circle is fixed by its radius (or diameter). Two circles are congruent if their radii are equal.

(b) A rectangle is fixed by its length and breadth (all angles are 90°). Two rectangles are congruent if their lengths are equal and their breadths are equal.

(a) The radius: circles with equal radii are congruent. (b) Length and breadth: rectangles with equal length and equal breadth are congruent.

Measure three lengths and one angle: the length of the long segment, the distance from its lower end to the point where the branch meets it, the length of the branch, and the angle between the branch and the segment. If all four are equal, the figures are congruent.

For both pairs shown, these measurements come out equal (within the thickness of the lines), so both pairs are congruent.

Compare the segment length, the position of the joining point, the branch length and the angle between them; in both pairs they match, so both pairs are congruent.

Yes. With AB = 6 cm and arcs of 4 cm and 8 cm, the arcs meet at E (above AB) and F (below). ΔABE and ΔABF are mirror images in the line AB (the construction above and below AB is identical), so they are congruent. So all triangles with the same three sides are congruent: the SSS condition.

Yes; ΔABE and ΔABF are mirror images in AB, so all triangles with the same sides are congruent (SSS).

Only if the triangle has some equal sides. For a scalene triangle, there is only one way: A on X, B on Y, C on Z (each side must land on the side of the same length). If the triangles were isosceles or equilateral, more ways would be possible.

For scalene triangles, no: the only correspondence is A↔X, B↔Y, C↔Z.

AB = CD, AD = CB and BD is common, so the triangles are congruent (SSS). Equal sides must overlap: AB on CD, AD on CB, BD on DB. So A ↔ C, B ↔ D, D ↔ B:

$$\Delta ABD\cong \Delta CDB$$

(written with the vertices in matching order: A–C, B–D, D–B).

A ↔ C, B ↔ D, D ↔ B: ΔABD ≅ ΔCDB.

The correspondence is H ↔ B, E ↔ I, N ↔ G. Any order of the vertices works as long as matching vertices stay in the same positions:

ΔHNE ≅ ΔBGI, ΔEHN ≅ ΔIBG, ΔENH ≅ ΔIGB, ΔNHE ≅ ΔGBI, ΔNEH ≅ ΔGIB (and of course ΔHEN ≅ ΔBIG).

ΔHNE ≅ ΔBGI, ΔEHN ≅ ΔIBG, ΔENH ≅ ΔIGB, ΔNHE ≅ ΔGBI, ΔNEH ≅ ΔGIB

All three sides are equal in pairs, so they are congruent (SSS). Matching the sides: R is where the 3.5 cm and 6 cm sides meet, and so is J; E is where 3.5 cm and 5 cm meet, and so is A; D and M join the 5 cm and 6 cm sides.

$$\Delta RED\cong \Delta JAM$$

Yes, by SSS: ΔRED ≅ ΔJAM.

In ΔABC and ΔADC: AB = AD, CB = CD and AC is common. So ΔABC ≅ ΔADC (SSS).

Corresponding angles are equal: ∠BAC = ∠DAC and ∠BCA = ∠DCA. So yes, AC divides both ∠BAD and ∠BCD into two equal parts.

ΔABC ≅ ΔADC (SSS); hence ∠BAC = ∠DAC and ∠BCA = ∠DCA, so AC bisects both angles.

The triangles DFE and DGE have DF = DG, FE = GE and the common side DE, so they are congruent by SSS, with D ↔ D, F ↔ G, E ↔ E:

$$\Delta DFE\cong \Delta DGE$$

But the statement "ΔDFE ≅ ΔGED" is not correct as written: it would match D with G and F with E, which are not corresponding vertices.

The triangles are congruent (SSS), but the correct statement is ΔDFE ≅ ΔDGE, not ΔDFE ≅ ΔGED.

No. Triangles with angles 30°, 70°, 80° can be of any size: they have the same shape but not necessarily the same size. So equal angles alone (AAA) do not guarantee congruence.

No; equal angles give the same shape but possibly different sizes.

Draw AB = 6 cm, make ∠A = 30° and mark C on the new arm with AC = 5 cm; join BC. Every student gets the same triangle: once the angle and the two arm lengths are fixed, the third side BC is fixed, so all such triangles are congruent (SAS).

Yes; the angle and the two arms fix the third side, so the triangles are congruent (SAS).

Not necessarily. Drawing PQ = 6 cm and a 30° line from P, an arc of radius 4 cm from Q cuts that line at two points R and S. ΔPQR and ΔPQS both fit the measurements but have different shapes. So the SSA condition does not guarantee congruence.

No; the 4 cm arc cuts the 30° line at two points, giving two different triangles (SSA fails).

Yes (ASA). Drawing the 5 cm side and the two angles at its ends, the arms always meet at the same point, so all such triangles are congruent: ΔABC ≅ ΔXYZ.

Yes, by ASA.

AO = OD, BO = OC and ∠AOB = ∠DOC (vertically opposite). So ΔAOB ≅ ΔDOC (SAS) and the corresponding sides AB and DC are equal.

AB = DC, because ΔAOB ≅ ΔDOC (SAS).

The 47° angle lies between the 7 cm and 5 cm sides in both triangles, so by SAS they are congruent, with B ↔ Z, A ↔ X, C ↔ Y:

$$\Delta ABC\cong \Delta XZY$$

Yes, SAS: ΔABC ≅ ΔXZY.

Since CD ∥ AB:

• ∠ODC = ∠OBA (alternate angles, transversal DB)
• ∠OCD = ∠OAB (alternate angles, transversal CA)
• and CD = AB (given).

So ΔODC ≅ ΔOBA (ASA). Hence also OD = OB, OC = OA (O is the midpoint of both AC and BD), and ∠DOC = ∠BOA.

∠ODC = ∠OBA and ∠OCD = ∠OAB (alternate angles); ΔODC ≅ ΔOBA (ASA), so OD = OB and OC = OA.

In ΔABC and ΔDBC: ∠ABC = ∠DBC, BC is common, ∠ACB = ∠DCB. So ΔABC ≅ ΔDBC (ASA), and hence ∠BAC = ∠BDC (corresponding angles).

(Even without congruence: $\angle BAC=180°-\angle ABC-\angle ACB=180°-\angle DBC-\angle DCB=\angle BDC$.)

ΔABC ≅ ΔDBC (ASA), so ∠BAC = ∠BDC; yes, they are congruent.

Adding the given equal angles: $\angle ABC=\angle ABD+\angle DBC=\angle DCA+\angle ACB=\angle DCB$.

So in ΔABC and ΔDCB: ∠ABC = ∠DCB, BC = CB (common), ∠ACB = ∠DBC. Hence ΔABC ≅ ΔDCB (ASA).

Equal parts: AB = DC, AC = DB, ∠BAC = ∠CDB, and ∠ABC = ∠DCB. (Also, if the diagonals meet at O, then OB = OC and OA = OD.)

ΔABC ≅ ΔDCB (ASA): AB = DC, AC = DB, ∠BAC = ∠CDB, ∠ABC = ∠DCB.

Yes, the arc of radius 5 cm meets the line l again below QR, at P′. But ΔP′QR is just the mirror image of ΔPQR in QR (QP′ = QP, both angles at Q are 90°), so it is congruent to ΔPQR. Hence all right triangles with the same hypotenuse and one other side are congruent (RHS).

Yes, but the lower triangle is the mirror image of ΔPQR in QR, so it is congruent (RHS holds).

Angles opposite equal sides are equal, so ∠B = ∠C and $\angle B+\angle C=180°-80°=100°$: ∠B = ∠C = 50°.

For an equilateral triangle all three angles are equal, so each is $180°÷3=60°$ (constructing one with sides of 5 cm and measuring confirms 60°).

∠B = ∠C = 50°; each angle of an equilateral triangle is 60°.

• Louvre pyramid: the glass is made of many identical small triangles (and diamonds made of two triangles); the four sloping faces are congruent triangles.
• Pyramid of Giza: its four sloping faces are congruent isosceles triangles.
• Dome design: the dome is covered by repeated congruent triangular panels in rings.
• Rangoli: the design repeats the same triangle shape around the centre, by turning and flipping.
• Howrah Bridge: the steel framework is made of many congruent triangles, which make it rigid and strong.

The pyramid faces, the glass panels, the dome panels, the repeated rangoli pieces and the bridge's steel triangles are congruent triangles.

• Vertices: A ↔ F, I ↔ L, R ↔ Y
• Sides: AI ↔ FL, IR ↔ LY, AR ↔ FY
• Angles: ∠A ↔ ∠F, ∠I ↔ ∠L, ∠R ↔ ∠Y

A–F, I–L, R–Y; AI = FL, IR = LY, AR = FY; ∠A = ∠F, ∠I = ∠L, ∠R = ∠Y.

(a) SSS: ΔABC ≅ ΔDEF.

(b) SAS (∠A is between AB and AC; ∠E is between EF and ED): A ↔ E, B ↔ F, C ↔ D, so ΔABC ≅ ΔEFD.

(c) RHS (right angles at B and D; hypotenuses AC and FE; sides AB and DF): A ↔ F, B ↔ D, C ↔ E, so ΔABC ≅ ΔFDE.

(d) AAS (two angles and the side AC, which is not between them): A ↔ D, B ↔ E, C ↔ F, so ΔABC ≅ ΔDEF.

(e) Not necessarily congruent. ∠B is not the angle between AB and AC (that would be ∠A), so this is the SSA case, which does not guarantee congruence.

(a) SSS: ΔABC ≅ ΔDEF (b) SAS: ΔABC ≅ ΔEFD (c) RHS: ΔABC ≅ ΔFDE (d) AAS: ΔABC ≅ ΔDEF (e) SSA, not necessarily congruent.

In ΔAOB and ΔDOC: OA = OD, OB = OC and ∠AOB = ∠DOC (vertically opposite). So ΔAOB ≅ ΔDOC (SAS), and the corresponding angles ∠OAB = ∠ODC.

These are alternate angles made by the transversal AD with the lines AB and CD. Since they are equal, AB ∥ CD.

ΔAOB ≅ ΔDOC (SAS), so ∠OAB = ∠ODC; these alternate angles are equal, hence AB ∥ CD.

AB = AD, BC = DC (sides of a square) and AC is common, so ΔABC ≅ ΔADC (SSS).

Also AB = CD, BC = DA and CA = AC, so ΔABC ≅ ΔCDA too. So one triangle fits on the other in two different ways (because ΔABC is isosceles: AB = BC).

• Two ways: any two congruent isosceles triangles, e.g. ΔPQR and ΔXYZ with PQ = PR = XY = XZ and QR = YZ: ΔPQR ≅ ΔXYZ and ΔPQR ≅ ΔXZY.
• Six ways: two equilateral triangles of the same side: every one of the 6 correspondences of vertices works (ΔABC ≅ ΔXYZ, ΔXZY, ΔYXZ, ΔYZX, ΔZXY, ΔZYX).

ΔABC ≅ ΔADC (SSS) and also ΔABC ≅ ΔCDA; congruent isosceles triangles match in 2 ways, equal equilateral triangles in 6 ways.

AB = AC (radii), so ΔABC is isosceles and ∠B = ∠C. Then $\angle B+\angle C=180°-120°=60°$, so ∠B = ∠C = 30°.

∠B = ∠C = 30°

Name the four unlabelled points inside: P (joined to U, R, O and Q), O (the point where many lines meet, joined to R, V, D, F, S, Q and P), Q (joined to U, P, O, S, K and A) and S (joined to O, Q, K, L, B and F). From the tick marks: CU = UA = CR = RV = VD = VO = UP = UQ; RO = AQ; KL = LB; SF = SB = FB. R, O, S lie on one straight line (44° + 46° + 90° = 180°).

Checks: at Q: 73° + 51° + 34° + 102° + 44° + 56° = 360°; at O: 44° + 46° + 90° + 64° + 26° + 34° + 56° = 360°.

E.g. ∠CUR = 45°, ∠AUQ = 68°, ∠UPQ = 73°, ∠VOR = 56°, ∠RPO = 102°, ∠OQS = 34°, ∠AKQ = 102°, ∠SKL = 30°, ∠OFS = 22°, ∠SOF = 64°, ∠VDO = 34°, ∠DOF = 26° (full list in the table).

There are 24 white squares, so each region has 4 squares. A computer search shows that only the L-shaped piece works. One way (each letter is one region):

All six regions are the same L shape (turned or flipped).

Six L-shaped pieces of 4 squares each, as in the grid above.