Question: Straight lines $PR$ and $QS$ cross at $O$. Given $PR = QS$, $PO = QO$ and $\angle PSQ = 32°$, show that triangles $POS$ and $QOR$ are congruent, and hence find $\angle PRQ$.

[diagram in original sheet: two crossing lines forming triangles POS and QOR on opposite sides of O]

Step 1. Collect three matching facts, each with a reason in brackets:

(S) $OS = OR$ (since $PR = QS$ and $PO = QO$ — subtract the equal parts)

(S) $OP = OQ$ (given)

(A) $\angle POS = \angle QOR$ (vertically opposite $\angle$s)

Step 2. Check the order: the angle sits between the two pairs of sides — that is Side-Angle-Side.

$$\therefore \triangle POS \equiv \triangle QOR \quad \text{(SAS)}$$

Step 3. Congruent triangles have equal corresponding angles.

$\angle PRQ$ (in $\triangle QOR$) corresponds to $\angle QSP$ (in $\triangle POS$):

$$\angle PRQ = \angle PSQ = 32°$$

⚠ Watch out: SAS needs the included angle — $\angle POS$ lies between $OP$ and $OS$; an angle elsewhere gives ASS, which proves nothing. And every statement needs its reason in brackets, or the proof marks are lost.

Question: $Q$ is the midpoint of the straight line $PR$. $SP$ is parallel to $TQ$, and $SQ$ is parallel to $TR$. Prove that $\triangle PSQ$ is congruent to $\triangle QTR$.

[diagram in original sheet: triangles PSQ and QTR side by side on the line PR]

Step 1. List the matching facts with reasons:

(S) $PQ = QR$ ($Q$ is the midpoint of $PR$)

(A) $\angle SPQ = \angle TQR$ (corresponding $\angle$s, $SP$ // $TQ$)

(A) $\angle SQP = \angle TRQ$ (corresponding $\angle$s, $SQ$ // $TR$)

Step 2. Check the pattern: the side $PQ$ lies between the two angles used — Angle-Side-Angle.

$$\therefore \triangle PSQ \equiv \triangle QTR \quad \text{(ASA)}$$

Step 3. State the congruence with the vertices in matching order: $P \leftrightarrow Q$, $S \leftrightarrow T$, $Q \leftrightarrow R$ — the order tells the reader which sides and angles correspond.

The five valid congruence tests are SSS, SAS, ASA, AAS and RHS.

⚠ Watch out: AAA and ASS are not congruence proofs — AAA only fixes the shape (that's similarity), and ASS allows two different triangles. Also write $\equiv$ for congruent, not $=$; and $\triangle PSQ \equiv \triangle QTR$ is wrong if the letters don't correspond pairwise.

Question: In $\triangle ABC$, the point $C$ lies on the line $BD$ with $BC = 4$ cm and $CD = 21$ cm, and $\angle BAC = \angle BDA$. Show that $\triangle ABC$ is similar to $\triangle DBA$, and find $AB$.

[diagram in original sheet: triangle ABD with C on BD, and AC drawn]

Step 1. Prove similarity with two angle pairs (AA):

$\angle BAC = \angle BDA$ (given)

$\angle ABC = \angle DBA$ (same angle — both triangles contain the angle at $B$)

$$\therefore \triangle ABC \sim \triangle DBA \quad \text{(AA)}$$

Step 2. Write the matching sides from the similarity statement — $A\leftrightarrow D$, $B\leftrightarrow B$, $C\leftrightarrow A$:

$$\dfrac{AB}{DB} = \dfrac{BC}{BA}$$

Step 3. Substitute $DB = BC + CD = 4 + 21 = 25$:

$$\dfrac{AB}{25} = \dfrac{4}{AB}$$

Step 4. Cross-multiply and solve:

$$AB^2 = 100 \;\Rightarrow\; AB = 10 \text{ cm}$$

⚠ Watch out: Read the matching sides straight off the letter order of $\triangle ABC \sim \triangle DBA$ — here $AB$ (in the small triangle) matches $DB$ (in the big one), and the shared side $BA$ appears in both ratios. Guessing pairs by eye is where these questions go wrong.

Question: Triangles $ABC$ and $DBA$ are similar, with corresponding sides $AB = 10$ cm and $DB = 25$ cm. Find the value of $\dfrac{\text{area of } \triangle ABC}{\text{area of } \triangle DBA}$.

Step 1. For similar figures, the ratio of areas is the ratio of corresponding lengths squared:

$$\dfrac{\text{area of } \triangle ABC}{\text{area of } \triangle DBA} = \left(\dfrac{AB}{DB}\right)^2$$

Step 2. Substitute the corresponding sides:

$$= \left(\dfrac{10}{25}\right)^2 = \left(\dfrac{2}{5}\right)^2$$

Step 3. Square it:

$$= \dfrac{4}{25}$$

Step 4. A useful follow-up trick: on this figure $\triangle DBA$ splits into $\triangle ABC$ and $\triangle ACD$, so in area units,

$$\triangle ABC = 4, \quad \triangle DBA = 25 \;\Rightarrow\; \triangle ACD = 25 - 4 = 21 \text{ units}$$

So, for example, if the area of $\triangle ABC$ is $14$ cm$^2$, then $\dfrac{14}{\text{area } \triangle ACD} = \dfrac{4}{21}$ gives area of $\triangle ACD = 73.5$ cm$^2$.

⚠ Watch out: Square the length ratio for areas — writing $\frac{10}{25}$ instead of $\left(\frac{10}{25}\right)^2$ is the single most common similarity error. And for a part of the figure, subtract first: the remaining piece is $25 - 4 = 21$ units, not $25$.

Question: $ABCD$ is a parallelogram. $S$ is a point on $AD$ with $SD : AS = 2 : 1$. The lines $SC$ and $BD$ meet at $T$. Find $DT : TB$, and hence the ratio of the area of $\triangle CTD$ to the area of $\triangle CBD$.

[diagram in original sheet: parallelogram ABCD with S on AD, SC and diagonal BD crossing at T]

Step 1. $\triangle STD \sim \triangle CTB$ (alternate $\angle$s twice, since $AD$ // $BC$, then AA).

Step 2. Get the length ratio of that similar pair. Since $SD:AS = 2:1$, $SD = \dfrac{2}{3}AD$, and $BC = AD$ (parallelogram):

$$\dfrac{SD}{CB} = \dfrac{2}{3} \;\Rightarrow\; DT : TB = 2 : 3$$

Step 3. Now the key idea: $\triangle CTD$ and $\triangle CBD$ have the same height — the perpendicular from $C$ to the line $DB$ serves both. So their areas are simply in the ratio of their bases:

$$\dfrac{\text{area } \triangle CTD}{\text{area } \triangle CBD} = \dfrac{DT}{DB} = \dfrac{2}{2+3} = \dfrac{2}{5}$$

⚠ Watch out: Same-height triangles compare by the plain base ratio — no squaring, because these two triangles are not similar to each other. Square the ratio only when comparing a similar pair (like $STD$ and $CTB$, whose area ratio is $\left(\frac{2}{3}\right)^2 = \frac{4}{9}$). Knowing which rule applies is the whole question.

Question: A shop sells two geometrically similar solid statues with heights in the ratio $2 : 5$. The circumference of the base of the smaller statue is $18$ cm; the mass of the larger is $400$ g; painting the larger costs $\usd{}15.50$. Find the larger circumference, the smaller mass, and the smaller painting cost.

Step 1. Sort each quantity by its dimension:

circumference → a length (ratio $2:5$); painted surface → an area (ratio $2^2:5^2$); mass → behaves like volume (ratio $2^3:5^3$).

Step 2. Circumference (length, scale up):

$$18 \times \dfrac{5}{2} = 45 \text{ cm}$$

Step 3. Mass (volume, scale down):

$$400 \times \left(\dfrac{2}{5}\right)^3 = 400 \times \dfrac{8}{125} = 25.6 \text{ g}$$

Step 4. Painting cost (proportional to surface area, scale down):

$$15.50 \times \left(\dfrac{2}{5}\right)^2 = 15.50 \times \dfrac{4}{25} = \usd{}2.48$$

⚠ Watch out: The exam skill is classifying the quantity before touching the ratio — circumference, height, radius scale as $k$; any area or cost-to-paint as $k^2$; volume, capacity and mass (same material) as $k^3$. Mass with a length ratio ($400 \times \frac{2}{5}$) is the classic wrong answer.

Question: Two similar flower pots have base areas $45$ cm$^2$ and $80$ cm$^2$. Find the ratio of their heights. The curved surface area of the smaller pot is $585$ cm$^2$ — find the larger pot's. The capacity of the larger pot is $9.6$ litres — find the smaller pot's capacity in cm$^3$.

Step 1. Go from area ratio back to length ratio with a square root:

$$\dfrac{h_1}{h_2} = \sqrt{\dfrac{45}{80}} = \sqrt{\dfrac{9}{16}} = \dfrac{3}{4}$$

Step 2. Curved surface areas are areas, so they share the $\dfrac{45}{80}$ ratio directly:

$$\dfrac{585}{A_2} = \dfrac{45}{80} \;\Rightarrow\; A_2 = \dfrac{585 \times 80}{45} = 1040 \text{ cm}^2$$

Step 3. Capacity is a volume, so cube the length ratio:

$$V_1 = \left(\dfrac{3}{4}\right)^3 \times 9.6 = \dfrac{27}{64} \times 9.6 = 4.05 \text{ litres}$$

Step 4. Convert the units: $1$ litre $= 1000$ cm$^3$:

$$V_1 = 4050 \text{ cm}^3$$

⚠ Watch out: Simplify $\frac{45}{80}$ to $\frac{9}{16}$ before square-rooting — the surd disappears. Working backwards: square root an area ratio, cube root a volume ratio. And $4.05$ litres is $4050$ cm$^3$, not $4.05$ cm$^3$.

Question: Triangle $A$ has sides $6$ cm, $8$ cm and $10$ cm. Triangle $B$ has sides $9$ cm, $12$ cm and $15$ cm. Triangle $C$ has sides $9$ cm, $12$ cm and $14$ cm. Is triangle $A$ similar to triangle $B$? Is it similar to triangle $C$?

Step 1. Two triangles are similar when the ratios of all three pairs of corresponding sides are equal — pair shortest with shortest, longest with longest.

Step 2. Test $A$ against $B$:

$$\dfrac{6}{9} = \dfrac{2}{3}, \qquad \dfrac{8}{12} = \dfrac{2}{3}, \qquad \dfrac{10}{15} = \dfrac{2}{3}$$

All three ratios equal $\dfrac{2}{3}$ → triangle $A$ is similar to triangle $B$ (three ratios equal).

Step 3. Test $A$ against $C$:

$$\dfrac{6}{9} = \dfrac{2}{3}, \qquad \dfrac{8}{12} = \dfrac{2}{3}, \qquad \dfrac{10}{14} = \dfrac{5}{7} \neq \dfrac{2}{3}$$

One ratio breaks → triangle $A$ is not similar to triangle $C$.

Step 4. The common ratio $\dfrac{2}{3}$ is the scale factor: every length of $A$ is $\dfrac{2}{3}$ of the matching length of $B$.

⚠ Watch out: All three ratios must match — two out of three is not enough. And always compare in a consistent order (small÷small, middle÷middle, large÷large); mixing them up can make similar triangles look non-similar.