Tracing works only for small figures. For a large figure, we instead record a few key measurements of the shape — such as the lengths of its arms and the angle between them — and then use those numbers to reconstruct the figure exactly, anywhere, at any scale of drawing tool.

For the “V” shaped symbol with corner points A, B, C, the natural measurements to try first are the two arm lengths, $AB$ and $BC$.

No. Knowing only the two lengths fixes how long each arm is, but says nothing about the angle at which they meet at $B$. We can swing the arm $BC$ around $B$ to any angle and both lengths stay $AB=4$ cm, $BC=8$ cm — producing many differently-shaped symbols.

AB and BC alone are NOT sufficient — the angle ∠ABC must also be fixed.

Yes. Measuring the angle $\angle ABC$ between the two arms, in addition to $AB$ and $BC$, removes the ambiguity seen above. The two arm lengths fix the “size” of each arm, and the included angle fixes exactly how they open — together the three measurements fully determine the shape and size of the figure.

Yes — this is a standard construction:

1. Draw the arm $BC = 8$ cm using a ruler.
2. At point $B$, use a protractor to draw a ray making $80^\circ$ with $BC$.
3. On this new ray, mark point $A$ such that $BA = 4$ cm.
4. Join $A$ to $B$ (already drawn) — the figure $A\!-\!B\!-\!C$ is the required symbol.

Any two people following these three measurements will draw identical (congruent) symbols.

No. As shown above, several non-congruent shapes share the same arm lengths $AB$ and $BC$ because the angle between the arms can differ. Equal arm lengths alone do not fix the angle, so we cannot conclude congruence from arm lengths alone.

We can only conclude congruence if the arm lengths and the included angle ∠ABC are all equal.

Exactly as with the signboard symbol: measure the two arm lengths and the angle between them for each figure, then compare. If both arm lengths and the included angle match between the two figures, the figures are congruent (this is the same idea used later for triangles as the SAS condition).

Method: compare (arm 1, arm 2, included angle) as a triple — equal triples ⇒ congruent figures.

Instead of tracing, they can take a small set of measurements from the frame (such as its three side lengths, or two sides and the angle between them) and then use these numbers to construct a new triangle on cardboard using a ruler, protractor and compass. If the right measurements are chosen, the new triangle will be an exact — congruent — copy of the original frame, however large it is.

Yes, Meera is correct. Unlike the two-armed “V” symbol (where one angle could vary freely), a closed triangle has three sides that constrain each other — once all three side lengths are fixed, there is essentially only one possible triangle shape (up to flipping), so the three side lengths alone are enough to guarantee an identical triangle. This is proved rigorously in the next question.

Yes. Construction (compass method):

1. Draw a line segment $AB = 6$ cm.
2. With centre $A$ and radius $4$ cm, draw an arc.
3. With centre $B$ and radius $8$ cm, draw another arc cutting the first.
4. The arcs intersect at two points, $E$ and $F$ (one above $AB$, one below). Either point can be joined to $A$ and $B$ to complete the triangle.

Compare the two triangles:

• $AE = AF = 4$ cm  (both are radii of the same arc centred at $A$)
• $BE = BF = 8$ cm  (both are radii of the same arc centred at $B$)
• $AB = AB$  (common side)

All three sides of $\triangle ABE$ equal all three sides of $\triangle ABF$. Also, line $AB$ acts as a line of symmetry — the construction above $AB$ mirrors the construction below it — so folding along $AB$ maps $E$ exactly onto $F$.

Yes — △ABE ≅ △ABF. Hence all triangles built from the same three sidelengths are congruent to each other (the SSS condition).

No. The tick marks show exactly which sides are equal: $AB = XY$, $BC = YZ$, $AC = XZ$. Superimposing the triangles so that equal sides lie on each other forces $A\!\to\!X$, $B\!\to\!Y$, $C\!\to\!Z$ — this is the only vertex correspondence under which every pair of equal sides overlaps. Any other pairing (say $A\to Y$) would try to lay an unequal pair of sides on top of each other, which fails.

Consider $\triangle ABD$ and $\triangle CDB$. Since $ABCD$ is a rectangle:

• $AB = CD$  (opposite sides of a rectangle)
• $AD = CB$  (opposite sides of a rectangle)
• $BD = DB$  (common diagonal)

All three sides of $\triangle ABD$ match all three sides of $\triangle CDB$, so by the SSS condition they are congruent. Matching the equal sides correctly ($AB \leftrightarrow CD$, $AD \leftrightarrow CB$, $BD\leftrightarrow DB$) gives the vertex correspondence $A\to C$, $B\to D$, $D\to B$.

△ABD ≅ △CDB

If we try $A\!\to\!C,\ B\!\to\!B,\ D\!\to\!D$, this would require side $AB$ of $\triangle ABD$ to overlap side $CB$ of $\triangle CDB$. But $AB$ (a rectangle’s length) need not equal $CB$ (a rectangle’s breadth) — so this pairing does not make the triangles overlap exactly, and is incorrect.

Correct correspondence: △ABD ≅ △CDB, i.e. A↔C, B↔D, D↔B.

No. Triangles with the same three angles can be different sizes — one a scaled-up (or down) copy of the other. They all have the same shape but not necessarily the same size.

Equal angles ⇒ same shape (similar triangles), but NOT necessarily congruent — size can still differ.

Construction: draw $AB=6$ cm; at $A$ draw a $30^\circ$ ray; on it mark $C$ so $AC=5$ cm; join $BC$. Because a ray from a fixed point at a fixed angle is unique, and a fixed length along it lands at one exact point, there is exactly one triangle possible with these three measurements. So everyone’s construction gives the same triangle.

Yes, all such triangles are congruent — this is the SAS (Side–Angle–Side) condition for congruence.

Yes — non-congruent triangles can exist. Construction: draw base $PQ = 6$ cm; at $P$ draw a ray at $30^\circ$; from $Q$, draw an arc of radius $4$ cm. This arc can cut the ray at two different points, $R$ and $S$, giving two triangles $\triangle PQR$ and $\triangle PQS$ that both satisfy the given measurements but are different in size/shape.

This is called the SSA (Side–Side–Angle) condition — it does NOT guarantee congruence.

Construction: draw $BC = 5$ cm; at $B$ draw a ray at $50^\circ$; at $C$ draw a ray at $30^\circ$; the two rays meet at exactly one point $A$. Since two fixed rays from two fixed points meet in only one place, the resulting triangle is unique.

Yes, congruent — △ABC ≅ △XYZ. No non-congruent triangle is possible. This is the ASA (Angle–Side–Angle) condition.

We know $AO=OD$ (O is midpoint of $AD$) and $BO=OC$ (O is midpoint of $BC$). Also, $\angle AOB = \angle DOC$ because they are vertically opposite angles at the crossing point $O$.

These three facts ($AO=OD$, $BO=OC$, included angle $\angle AOB=\angle DOC$) satisfy the SAS condition, so $\triangle AOB \cong \triangle DOC$.

Since AB and DC are corresponding sides of congruent triangles, AB = DC. (Also ∠OAB = ∠ODC and ∠OBA = ∠OCD, as corresponding angles.)

Step 1 — find the third angle: Angle sum of a triangle is $180^\circ$:

$$\angle B = 180^\circ – \angle A – \angle C = 180^\circ – 35^\circ – 75^\circ = 70^\circ$$

Similarly, $\angle Y = 180^\circ – 35^\circ – 75^\circ = 70^\circ$. So $\angle B = \angle Y = 70^\circ$.

Step 2 — apply ASA: Now we know $\angle B=\angle Y=70^\circ$, side $BC=YZ=4$ cm (the side included between $\angle B$ and $\angle C$), and $\angle C=\angle Z=75^\circ$. This satisfies the ASA condition.

Yes, △ABC ≅ △XYZ. Whenever two angles and any side are equal (even a non-included side), the third angle is forced to be equal too — turning it into an ASA case. This special case is called AAS (Angle-Angle-Side), and it always guarantees congruence.

Construction: draw $QR = 4$ cm; draw a perpendicular line $l$ at $Q$; from $R$, cut an arc of radius $5$ cm to meet $l$ at $P$. This determines $\triangle PQR$. If we also try extending line $l$ downward from $Q$, the arc from $R$ meets that extension too — but the triangle formed there is simply the mirror image of $\triangle PQR$, which is still congruent to it (just flipped).

So there is essentially only one triangle (up to a flip). The equal parts are: the right angle, the hypotenuse, and one leg — this special SSA case is guaranteed to work.

Yes, △ABC ≅ △XYZ — this is the RHS (Right angle–Hypotenuse–Side) condition.

Let $AD \perp BC$. Compare $\triangle ADB$ and $\triangle ADC$:

• $AB = AC$ (given — hypotenuse of each right triangle)
• $\angle ADB = \angle ADC = 90^\circ$ (by construction)
• $AD = AD$ (common side)

This satisfies the RHS condition, so $\triangle ADB \cong \triangle ADC$, which gives $\angle B = \angle C$ (corresponding angles of congruent triangles).

Finding the values: using the angle sum property, $\angle A + \angle B + \angle C = 180^\circ$. Since $\angle B=\angle C$:

$$80^\circ + 2\angle B = 180^\circ \ \Rightarrow\ \angle B = \frac{100^\circ}{2} = 50^\circ$$

∠B = ∠C = 50°. In general: angles opposite equal sides of a triangle are equal.

In an equilateral triangle $\triangle ABC$, all three sides are equal: $AB=BC=CA$. Using the fact that angles opposite equal sides are equal (proved above):

• Since $AB = AC$: $\angle B = \angle C$
• Since $AB = BC$: $\angle A = \angle C$

Combining both, $\angle A = \angle B = \angle C$. Using the angle sum property:

$$3 \times \angle A = 180^\circ \ \Rightarrow\ \angle A = 60^\circ$$

All three angles of an equilateral triangle equal 60° each.

• Louvre Museum Pyramid: its glass surface is built from many identically-sized triangular glass panels — congruent triangles tiled together give the structure its strength and symmetry.
• Egyptian Pyramid of Giza: the four sloping outer faces are (very nearly) congruent isosceles triangles meeting at the apex.
• Dome design: the geodesic dome is made of a network of congruent (or a few repeated sets of congruent) triangular frames — this evenly distributes load over the curved surface.
• Rangoli design: the eight-pointed star pattern is built from a set of congruent triangles arranged symmetrically around the centre, repeated by rotation.
• Howrah Bridge (Rabindra Setu): the steel truss is a repeating lattice of congruent triangular units — triangles keep their shape under load, so repeating congruent triangles makes the bridge rigid and strong.

To check, we trace one figure on tracing paper and superimpose it on the other (rotating/flipping as needed), or measure both arm lengths and the included angle of each and compare the triples.

From the figure, both shapes have the same pair of arm lengths and the same angle between them — so they superimpose exactly. They are congruent.

Method: trace one shape from each pair and try to place it exactly over the other (rotating or flipping is allowed).

• Cloud outlines — same size and same bumps, just possibly mirrored → congruent.
• Starburst shapes — same number of points and same spread → congruent.
• Droplet shapes — look similar but one is visibly rounder/larger than the other → not congruent (same shape family, different size).
• Petal/leaf pairs — check tip-to-tip length and curve; if identical when flipped → congruent.

Circle the cloud pair, the starburst pair, and the petal pair as congruent (matching size and shape under tracing).

The correspondence fixed by $\triangle HEN \cong \triangle BIG$ is: $H\leftrightarrow B$, $E\leftrightarrow I$, $N \leftrightarrow G$. Any way of writing the two names — as long as the same three positions in both names keep the same pairing — is a valid restatement. There are $3! = 6$ such orderings in total; one is given, so the other five are:

• $\triangle HNE \cong \triangle BGI$
• $\triangle EHN \cong \triangle IBG$
• $\triangle ENH \cong \triangle IGB$
• $\triangle NHE \cong \triangle GBI$
• $\triangle NEH \cong \triangle GIB$

Compare corresponding sides:

• $RE = JA = 3.5$ cm
• $ED = AM = 5$ cm
• $RD = JM = 6$ cm

All three sides match — this satisfies the SSS condition. Matching equal sides gives the correspondence $R\leftrightarrow J$, $E\leftrightarrow A$, $D\leftrightarrow M$.

Yes, congruent: △RED ≅ △JAM (by SSS).

Consider $\triangle ABC$ and $\triangle ADC$:

• $AB = AD$ (given)
• $CB = CD$ (given)
• $AC = AC$ (common side)

This satisfies SSS, so $\triangle ABC \cong \triangle ADC$.

Since corresponding angles of congruent triangles are equal: $\angle BAC = \angle DAC$ and $\angle BCA = \angle DCA$.

Yes — AC bisects both ∠BAD and ∠BCD, because △ABC ≅ △ADC.

Compare $\triangle DFE$ and $\triangle DGE$:

• $DF = DG$ (given)
• $FE = GE$ (given)
• $DE = DE$ (common side)

Yes — by SSS, △DFE ≅ △DGE.

In $\triangle ABC$, the $47^\circ$ angle at $B$ lies between the two given sides $AB$ and $BC$. In $\triangle XYZ$, the $47^\circ$ angle at $Z$ lies between the two given sides $XZ$ and $YZ$ — so in both triangles the angle is the included angle.

Matching $AB=XZ=7$, $BC=YZ=5$, included $\angle B=\angle Z=47^\circ$ satisfies SAS.

Yes, congruent: △ABC ≅ △XZY (by SAS).

$AC$ and $BD$ cross at a point, say $O$, with $A,B$ on one side and $C,D$ on the other. Since $AB \parallel CD$:

• $\angle ABO = \angle CDO$ (alternate angles, transversal $BD$)
• $\angle BAO = \angle DCO$ (alternate angles, transversal $AC$)
• $AB = CD$ (given)

These satisfy ASA, so $\triangle ABO \cong \triangle CDO$. Consequently $AO = CO$ and $BO = DO$ as well.

Yes, congruent: △ABO ≅ △CDO (by ASA); this also gives AO = CO and BO = DO.

Consider $\triangle ABC$ and $\triangle DBC$:

• $\angle ABC = \angle DBC$ (given)
• $BC = BC$ (common side)
• $\angle ACB = \angle DCB$ (given)

This satisfies ASA, so $\triangle ABC \cong \triangle DBC$. Since $\angle BAC$ and $\angle BDC$ are the remaining (corresponding) angles of these congruent triangles, they must be equal.

∠BAC = ∠BDC, and △ABC ≅ △DBC (by ASA).

Add $\angle DBC\, (=\angle ACB)$ to both sides of the given equality $\angle ABD = \angle DCA$:

$$\angle ABD + \angle DBC \;=\; \angle DCA + \angle ACB$$

The left side is $\angle ABC$ and the right side is $\angle DCB$, so $\angle ABC = \angle DCB$.

Now compare $\triangle ABC$ and $\triangle DCB$:

• $\angle ACB = \angle DBC$ (given)
• $BC = CB$ (common side)
• $\angle ABC = \angle DCB$ (just shown)

By ASA, △ABC ≅ △DCB. This also gives AB = DC, AC = DB, and ∠BAC = ∠CDB.

$\angle B = 180^\circ-35^\circ-75^\circ = 70^\circ$, and likewise $\angle Y=70^\circ$. Now $\angle B=\angle Y$, side $BC=YZ$ (included between $B$ and $C$), and $\angle C=\angle Z$ — this is ASA.

Yes — △ABC ≅ △XYZ, confirming the AAS condition (two angles + a non-included side) always works, since the third angle becomes fixed.

Corresponding vertices: $A\leftrightarrow F$, $I\leftrightarrow L$, $R\leftrightarrow Y$

Corresponding sides: $AI \leftrightarrow FL$, $IR \leftrightarrow LY$, $RA \leftrightarrow YF$

Corresponding angles: $\angle A \leftrightarrow \angle F$, $\angle I \leftrightarrow \angle L$, $\angle R \leftrightarrow \angle Y$

Consider $\triangle AOB$ and $\triangle DOC$:

• $OA = OD$ (given)
• $\angle AOB = \angle DOC$ (vertically opposite angles at $O$)
• $OB = OC$ (given)

By SAS, $\triangle AOB \cong \triangle DOC$. So $\angle OAB = \angle ODC$ (corresponding angles).

But $\angle OAB$ and $\angle ODC$ are alternate angles formed by transversal $AD$ cutting lines $AB$ and $DC$. Equal alternate angles mean the lines are parallel.

Hence AB ∥ CD.

Part 1: In square $ABCD$: $AB=AD$ (sides of square), $BC=DC$ (sides of square), $AC=AC$ (common diagonal) → SSS → $\triangle ABC \cong \triangle ADC$.

Part 2: Because every side of a square is equal ($AB=BC=CD=DA$), the same diagonal also gives $AB=CD$, $BC=DA$, $AC=CA$ — so SSS is satisfied again with a different vertex matching.

Yes — △ABC ≅ △CDA as well (a second, different correspondence), because a square’s extra symmetry allows two valid congruences at once.

More two-way examples: a rectangle split by a diagonal (as in IT-13 above); a rhombus split by either diagonal.

A six-way example: Take any equilateral triangle $\triangle ABC$ and compare it to itself. Since all its sides and all its angles are equal, every permutation of its vertices gives a valid congruence:

$$\triangle ABC \cong \triangle BCA \cong \triangle CAB \cong \triangle ACB \cong \triangle CBA \cong \triangle BAC$$

— six different, equally valid ways, because full symmetry removes any distinction between the vertices.

Since $A$ is the centre, $AB$ and $AC$ are both radii, so $AB = AC$ — $\triangle ABC$ is isosceles, and angles opposite the equal sides are equal: $\angle B = \angle C$.

By the angle sum property:

$$\angle A + \angle B + \angle C = 180^\circ \ \Rightarrow\ 120^\circ + 2\angle B = 180^\circ \ \Rightarrow\ \angle B = \frac{60^\circ}{2} = 30^\circ$$

∠B = ∠C = 30°

This puzzle is solved purely by repeatedly applying three tools, moving from the labelled angles outward:

1. Isosceles-triangle rule: wherever two sides carry the same tick marks (single “|”, double “||”, etc.), the angles opposite those sides are equal.
2. Angle sum of a triangle = $180^\circ$, used inside every small triangle formed by the crossing lines.
3. Linear pair / right angle at the rectangle’s corners ($90^\circ$ at $A, B, C, D$) and vertically opposite angles at every interior crossing point.

Worked example (corner $A$): The rectangle’s corner angle at $A$ is $90^\circ$. It is split by segment $AU$ into two labelled parts, $56^\circ$ and $34^\circ$:

$$56^\circ + 34^\circ = 90^\circ \ \checkmark$$

which confirms the two given readings are consistent with the right angle at $A$.

Worked example (triangle $AUK$): the tick marks show $AU$ and $UK$ are the equal-marked sides, so the triangle is isosceles with base angles at $A$ and $K$ equal. Using the angle sum $180^\circ$ together with the marked $34^\circ$ at $U$ lets the remaining base angles be pinned down consistently with the other labels around $U$ ($44^\circ$, $46^\circ$) and around $K$ ($44^\circ$, $30^\circ$) — each interior triangle’s three angles are checked to add to $180^\circ$ as a consistency test.

Worked example (top region near $R$ and $V$): $\angle R = 34^\circ$ and $\angle V = 68^\circ$ sit in a triangle together with the $44^\circ$ and $46^\circ$ angles at the central crossing point; since $34^\circ+68^\circ+… $ chase around the point where several lines cross, opposite angles are equal (vertical angles), e.g. the $90^\circ$ mark at the centre pairs with another $90^\circ$ vertically opposite, and $56^\circ$ pairs with $56^\circ$.

General method to fill in every blank vertex in such a figure:

• At any point where lines cross, opposite angles are equal (vertical angles) and angles on a straight line add to $180^\circ$.
• In any small triangle, once two angles are known (from tick marks or previous steps), the third is $180^\circ$ minus their sum.
• Tick-marked equal sides mean the triangle is isosceles, so its base angles are equal — use this to transfer a known angle to its “twin” angle across the triangle.

Applying these three rules triangle-by-triangle across the figure (starting from the labelled 34°, 56°, 90°, 44°, 46°, 30°, 68°, 98° and working inward/outward) fills in every remaining angle — each unmarked vertex angle is found as 180° minus the sum of its two known neighbours in that triangle, or as the vertical/alternate angle of an already-found value.