We often move from congruence — figures that are identical in shape and size — to the more nuanced and powerful idea of similarity. Similarity is the mathematics behind every map, every photograph, and every engineering blueprint. It is the very logic Thales used to measure the height of the Egyptian pyramids using nothing but shadow lengths.

In this guide, each section contains five problems arranged from foundational to challenging. Every solution is worked step by step, and the first problem of each set includes a Verification Check so you learn to confirm your own work. Click Show Solution to reveal the full working.

01

Similar Figures & Polygons

Two polygons are similar when corresponding angles are equal and corresponding sides are in the same ratio. Both conditions must hold simultaneously.

Core Definition

Two polygons of the same number of sides are similar if:

  • All corresponding angles are equal, and
  • All corresponding sides are in the same ratio (the scale factor).
Key insight: All congruent figures are similar, but similar figures need not be congruent. Congruence is the special case where the scale factor equals exactly \(1\).
P1 Are all equilateral triangles similar? Foundational
Prove geometrically whether any two equilateral triangles must be similar to each other.

Step-by-Step Solution

1
Recall the angle sum of a triangle: the interior angles of any triangle sum to \(180°\).
2
In an equilateral triangle, all three sides are equal, so all three angles are equal. Therefore each angle is \(\dfrac{180°}{3} = 60°\). This is true for every equilateral triangle, regardless of its side length.
3
Check Condition 1 (Angles): Any two equilateral triangles have angles \(60°, 60°, 60°\). All corresponding angles are equal. ✓
4
Check Condition 2 (Sides): Let the first triangle have side \(a\) and the second have side \(b\). The ratio of every pair of corresponding sides is \(\dfrac{a}{b}\), which is a single constant. The sides are proportional. ✓
5
Since both conditions are satisfied, any two equilateral triangles are similar.
✓ Yes — all equilateral triangles are similar.
✦ Verification Check Take two specific triangles: side 3 cm and side 5 cm. Angles: both \(60°, 60°, 60°\) — equal. Ratio of sides: \(\dfrac{3}{5}\) for every pair — constant. Both conditions met. Confirmed! ✓
P2 Finding the scale factor Basic
Quadrilateral \(ABCD\) has sides \(2, 4, 6, 8\) units. Quadrilateral \(PQRS\) has corresponding sides \(3, 6, 9, 12\) units. Calculate the scale factor and verify similarity (assume equal angles).

Step-by-Step Solution

1
Identify each corresponding pair: \(2 \to 3\), \(4 \to 6\), \(6 \to 9\), \(8 \to 12\).
2
Compute each ratio:
\(\dfrac{3}{2} = 1.5, \quad \dfrac{6}{4} = 1.5, \quad \dfrac{9}{6} = 1.5, \quad \dfrac{12}{8} = 1.5\)
3
All four ratios equal \(1.5\). The sides are proportional, and with equal angles (given), both similarity conditions hold.
4
Express as a fraction: \(\dfrac{3}{2}\).
✓ Scale factor \(= \dfrac{3}{2} = 1.5\). The quadrilaterals are similar.
P3 Using the scale factor to find an unknown side Intermediate
Polygon \(X\) is similar to polygon \(Y\) with a scale factor of \(2.5\) (meaning \(Y\) is larger). A side in polygon \(X\) measures \(10\text{ cm}\). Find the corresponding side in polygon \(Y\).

Step-by-Step Solution

1
The scale factor is defined as: \(\text{Scale Factor} = \dfrac{\text{Side in } Y}{\text{Side in } X}\).
2
Substitute the known values:
\(2.5 = \dfrac{\text{Side}_Y}{10}\)
3
Multiply both sides by \(10\):
\(\text{Side}_Y = 2.5 \times 10 = 25\text{ cm}\)
✓ The corresponding side in polygon \(Y\) is \(25\text{ cm}\).
P4 A tricky two-condition check Challenging
A square and a rhombus both have all side lengths equal to \(5\text{ cm}\). A student claims they must be similar because their sides are proportional (ratio \(= 1\)). Evaluate this claim carefully.

Step-by-Step Solution

1
Check Condition 2 (Sides): Both have four equal sides of 5 cm. The ratio is \(\dfrac{5}{5} = 1\). Sides are proportional. ✓
2
Check Condition 1 (Angles): Every interior angle of a square is exactly \(90°\). A rhombus has equal sides but its angles are not required to be \(90°\). For example, a rhombus could have angles \(60°, 120°, 60°, 120°\).
3
Comparing: Square has \(90°, 90°, 90°, 90°\). Rhombus has, say, \(60°, 120°, 60°, 120°\). These are NOT equal — Condition 1 fails. ✗
4
The student's claim is incorrect. Both conditions must hold simultaneously. Passing only one is insufficient for similarity.
✗ The claim is false. Corresponding angles are not equal, so the square and rhombus are NOT similar.
P5 Photography enlargement problem Advanced
A photographer enlarges a \(35\text{ mm}\) film image to a \(45\text{ mm}\) print. A geometric figure on the original film has a base of \(7\text{ mm}\). Find the base in the enlarged print and identify the Representative Fraction.

Step-by-Step Solution

1
The scale factor (enlargement ratio) is:
\(\text{Scale Factor} = \dfrac{\text{Enlarged size}}{\text{Original size}} = \dfrac{45}{35} = \dfrac{9}{7}\)
2
The Representative Fraction is \(\dfrac{9}{7}\) (each unit in the original corresponds to \(\dfrac{9}{7}\) units in the enlarged image).
3
Apply this scale factor to the figure's base:
\(\text{Enlarged base} = 7\text{ mm} \times \dfrac{9}{7} = \dfrac{63}{7} = 9\text{ mm}\)
4
Note that the angles in the enlarged photograph remain identical — only lengths change, not directions. This is the geometric essence of similarity.
✓ Enlarged base \(= 9\text{ mm}\). Representative Fraction \(= \dfrac{9}{7}\).
Before you leave, practice these: Quiz | Question Paper | Play & Study game.

✅ Play and Study ✅ Quiz ✅ Question Paper

02

The Basic Proportionality Theorem

Thales' great insight: a line parallel to one side of a triangle divides the other two sides in the same ratio — and the converse is equally powerful.

Theorems 6.1 & 6.2 (Thales / BPT)

Theorem 6.1 (BPT): If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then the other two sides are divided in the same ratio.

If \(DE \parallel BC\) in \(\triangle ABC\) with \(D\) on \(AB\) and \(E\) on \(AC\), then: \(\quad\dfrac{AD}{DB} = \dfrac{AE}{EC}\)

Theorem 6.2 (Converse): If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.

P1 Direct application of BPT Foundational
In \(\triangle ABC\), line \(DE \parallel BC\) with \(D\) on \(AB\) and \(E\) on \(AC\). Given \(AD = 1.5\text{ cm}\), \(DB = 3\text{ cm}\), and \(AE = 1\text{ cm}\), find \(EC\).

Step-by-Step Solution

1
Since \(DE \parallel BC\), by Theorem 6.1 (BPT):
\(\dfrac{AD}{DB} = \dfrac{AE}{EC}\)
2
Substitute the known values:
\(\dfrac{1.5}{3} = \dfrac{1}{EC}\)
3
Simplify the left side: \(\dfrac{1.5}{3} = 0.5 = \dfrac{1}{2}\).
4
Solve: \(\dfrac{1}{2} = \dfrac{1}{EC}\), so cross-multiplying: \(EC = 2\text{ cm}\).
✓ \(EC = 2\text{ cm}\)
✦ Verification Check Check: \(\dfrac{AD}{DB} = \dfrac{1.5}{3} = 0.5\) and \(\dfrac{AE}{EC} = \dfrac{1}{2} = 0.5\). Both ratios equal \(0.5\). The proportion holds — answer confirmed! ✓
P2 Testing parallelism using the converse Basic
In \(\triangle PQR\), points \(E\) and \(F\) lie on \(PQ\) and \(PR\). Given \(PE = 3.9\text{ cm}\), \(EQ = 3\text{ cm}\), \(PF = 3.6\text{ cm}\), and \(FR = 2.4\text{ cm}\), determine whether \(EF \parallel QR\).

Step-by-Step Solution

1
By Theorem 6.2 (Converse of BPT), \(EF \parallel QR\) if and only if \(\dfrac{PE}{EQ} = \dfrac{PF}{FR}\).
2
Calculate the first ratio:
\(\dfrac{PE}{EQ} = \dfrac{3.9}{3} = 1.3\)
3
Calculate the second ratio:
\(\dfrac{PF}{FR} = \dfrac{3.6}{2.4} = 1.5\)
4
Compare: \(1.3 \neq 1.5\). The ratios are not equal.
✗ \(EF\) is not parallel to \(QR\).
P3 Solving an algebraic BPT equation Intermediate
In a triangle, a line parallel to the base divides the two sides into segments of \(x\) and \((x+2)\) on the left, and \((x+3)\) and \((x+9)\) on the right. Find the value of \(x\).

Step-by-Step Solution

1
Since the line is parallel to the base, by BPT:
\(\dfrac{x}{x+2} = \dfrac{x+3}{x+9}\)
2
Cross-multiply:
\(x(x + 9) = (x + 2)(x + 3)\)
3
Expand the left side:
\(x^2 + 9x\)
4
Expand the right side step by step:
\((x+2)(x+3) = x^2 + 3x + 2x + 6 = x^2 + 5x + 6\)
5
Set both sides equal and cancel \(x^2\):
\(x^2 + 9x = x^2 + 5x + 6 \implies 9x = 5x + 6 \implies 4x = 6\)
6
Solve:
\(x = \dfrac{6}{4} = 1.5\)
✓ \(x = 1.5\)
P4 BPT in a trapezium (construction required) Challenging
\(ABCD\) is a trapezium where \(AB \parallel DC\). A line \(EF\) is drawn parallel to \(AB\), with \(E\) on \(AD\) and \(F\) on \(BC\). If \(AE = 4\), \(ED = 6\), and \(FC = 9\), find \(BF\).

Step-by-Step Solution

1
Construction: Draw diagonal \(AC\) to intersect \(EF\) at point \(G\). (This is the standard construction for this type of problem.)
2
Since \(AB \parallel DC\) and \(EF \parallel AB\), we get \(EF \parallel DC\). Now in \(\triangle ADC\), \(EG \parallel DC\). By BPT:
\(\dfrac{AE}{ED} = \dfrac{AG}{GC} \implies \dfrac{4}{6} = \dfrac{AG}{GC} \implies \dfrac{AG}{GC} = \dfrac{2}{3}\)
3
In \(\triangle CAB\), \(GF \parallel AB\). By BPT applied to this triangle:
\(\dfrac{AG}{GC} = \dfrac{BF}{FC}\)
4
Substitute the known ratio and value:
\(\dfrac{2}{3} = \dfrac{BF}{9}\)
5
Solve for \(BF\):
\(BF = \dfrac{2 \times 9}{3} = 6\)
✓ \(BF = 6\)
P5 Two-step BPT proof Advanced
In \(\triangle ABC\), \(DE \parallel AC\) and \(DF \parallel AE\), where \(F\) lies on \(BE\) and \(D\) lies on \(BA\). Prove that \(\dfrac{BF}{FE} = \dfrac{BE}{EC}\).

Step-by-Step Solution

1
In \(\triangle ABC\), since \(DE \parallel AC\), apply BPT:
\(\dfrac{BD}{DA} = \dfrac{BE}{EC} \quad \cdots (1)\)
2
In \(\triangle ABE\), since \(DF \parallel AE\), apply BPT:
\(\dfrac{BD}{DA} = \dfrac{BF}{FE} \quad \cdots (2)\)
3
Both equations (1) and (2) share the same left-hand side \(\dfrac{BD}{DA}\). By transitivity (if \(a = b\) and \(a = c\), then \(b = c\)):
\(\dfrac{BF}{FE} = \dfrac{BE}{EC}\)
4
This completes the proof. Notice how two applications of BPT — one in each sub-triangle — link through a shared intermediate ratio. This is the elegance of the theorem.
✓ Proven: \(\dfrac{BF}{FE} = \dfrac{BE}{EC}\)
03

Criteria for Similarity — AAA, SSS, SAS

We need not verify all six parts. Three elegant criteria let us confirm similarity efficiently.

The Three Criteria

  • AAA (AA): If two angles of one triangle equal two angles of another, they are similar. (The third angle follows from the angle sum property.)
  • SSS: If all three pairs of corresponding sides are in the same ratio, the triangles are similar.
  • SAS: If one angle is equal and the sides including that angle are proportional, the triangles are similar.
In triangles, unlike polygons, you only need ONE condition (angles OR sides ratio) — because in triangles, each condition implies the other (Theorems 6.3 and 6.4).
P1 AA criterion identification Foundational
\(\triangle ABC\) has \(\angle A = 80°\) and \(\angle B = 60°\). \(\triangle DEF\) has \(\angle D = 80°\) and \(\angle E = 60°\). Are the triangles similar? State the criterion.

Step-by-Step Solution

1
Find the third angle in each triangle. In \(\triangle ABC\): \(\angle C = 180° - 80° - 60° = 40°\). In \(\triangle DEF\): \(\angle F = 180° - 80° - 60° = 40°\).
2
List corresponding angles: \(\angle A = \angle D = 80°\), \(\angle B = \angle E = 60°\), \(\angle C = \angle F = 40°\). All three pairs are equal.
3
Two pairs of corresponding angles are equal — this is sufficient by the AA criterion (the third is automatic).
4
Therefore \(\triangle ABC \sim \triangle DEF\).
✓ Yes — \(\triangle ABC \sim \triangle DEF\) by the AA criterion.
✦ Verification Check Check: We confirmed \(\angle A = \angle D\), \(\angle B = \angle E\). Since two angles match, by the angle sum property the third must also match: \(40° = 40°\). ✓ AA criterion confirmed.
P2 SSS criterion to find a missing angle Basic
\(\triangle ABC\) has sides \(3.8, 6, 3\sqrt{3}\) with \(\angle A = 80°\), \(\angle B = 60°\). \(\triangle RQP\) has sides \(7.6, 12, 6\sqrt{3}\). Find \(\angle P\).

Step-by-Step Solution

1
Check each ratio of corresponding sides:
\(\dfrac{AB}{RQ} = \dfrac{3.8}{7.6} = \dfrac{1}{2}, \quad \dfrac{BC}{QP} = \dfrac{6}{12} = \dfrac{1}{2}, \quad \dfrac{CA}{PR} = \dfrac{3\sqrt{3}}{6\sqrt{3}} = \dfrac{1}{2}\)
2
All three ratios equal \(\dfrac{1}{2}\). By the SSS similarity criterion, \(\triangle ABC \sim \triangle RQP\).
3
In similar triangles, corresponding angles are equal. The vertex correspondence is \(A \leftrightarrow R\), \(B \leftrightarrow Q\), \(C \leftrightarrow P\). Therefore \(\angle C = \angle P\).
4
Find \(\angle C\) using the angle sum property:
\(\angle C = 180° - \angle A - \angle B = 180° - 80° - 60° = 40°\)
✓ \(\angle P = 40°\)
P3 SAS criterion with missing side Intermediate
In \(\triangle ABC\) and \(\triangle DEF\), \(\dfrac{AB}{DE} = \dfrac{AC}{DF} = \dfrac{2}{3}\) and \(\angle A = \angle D = 50°\). If \(BC = 10\text{ cm}\), find \(EF\).

Step-by-Step Solution

1
The two sides forming the included angle \(\angle A = \angle D\) are proportional (ratio \(\dfrac{2}{3}\)). This satisfies the SAS criterion.
2
Therefore, \(\triangle ABC \sim \triangle DEF\) by SAS similarity.
3
In similar triangles, all corresponding sides share the same ratio. So:
\(\dfrac{BC}{EF} = \dfrac{2}{3}\)
4
Substitute \(BC = 10\):
\(\dfrac{10}{EF} = \dfrac{2}{3} \implies 2 \times EF = 30 \implies EF = 15\text{ cm}\)
✓ \(EF = 15\text{ cm}\)
P4 Parallel lines and AAA similarity Challenging
Given \(PQ \parallel RS\) and line segments \(PR\) and \(QS\) intersecting at \(O\), prove that \(\triangle POQ \sim \triangle SOR\).

Step-by-Step Solution

1
Since \(PQ \parallel RS\), lines \(PR\) and \(QS\) act as transversals. By the Alternate Interior Angles theorem:
\(\angle QPO = \angle RSO \quad \text{(alternate angles, transversal } PR\text{)}\)
2
Similarly, with transversal \(QS\):
\(\angle PQO = \angle SRO \quad \text{(alternate angles, transversal } QS\text{)}\)
3
At point \(O\), \(\angle POQ = \angle SOR\) (vertically opposite angles).
4
Three pairs of corresponding angles are equal. By the AAA criterion: \(\triangle POQ \sim \triangle SOR\).
✓ \(\triangle POQ \sim \triangle SOR\) by the AAA criterion.
P5 The shadow and lamp-post problem Advanced
A girl of height \(90\text{ cm}\) walks away from the base of a lamp-post (height \(3.6\text{ m}\)) at \(1.2\text{ m/s}\). Calculate the length of her shadow after \(4\) seconds.

Step-by-Step Solution

1
Convert units: girl's height \(= 90\text{ cm} = 0.9\text{ m}\). Distance walked in 4 s: \(BD = 1.2 \times 4 = 4.8\text{ m}\). Let shadow length \(= x\) metres.
2
Set up the geometry: the lamp-post (\(AB\)) and girl (\(CD\)) are both vertical, making right angles with the ground. The tip of the shadow (\(E\)) creates \(\triangle ABE\) and \(\triangle CDE\).
3
Establish similarity: in \(\triangle ABE\) and \(\triangle CDE\), \(\angle B = \angle D = 90°\) and they share \(\angle E\). By AA: \(\triangle ABE \sim \triangle CDE\).
4
Write the proportion:
\(\dfrac{BE}{DE} = \dfrac{AB}{CD} \implies \dfrac{4.8 + x}{x} = \dfrac{3.6}{0.9}\)
5
Simplify the right side: \(\dfrac{3.6}{0.9} = 4\). So:
\(\dfrac{4.8 + x}{x} = 4 \implies 4.8 + x = 4x \implies 4.8 = 3x \implies x = 1.6\text{ m}\)
✓ The shadow length after 4 seconds is \(1.6\text{ m}\).
04

Challenge Problems — Mixed Concepts

These problems require you to chain multiple theorems and choose the right tool. This is where real mastery is built.

Strategy for Hard Problems

When faced with a complex problem: (1) identify what type of relationship is involved (ratio of sides, parallel lines, angles), (2) choose the most appropriate theorem (BPT, AA, SSS, or SAS), (3) introduce constructions (auxiliary lines) when needed, and (4) work algebraically with precision.

P1 Vertical pole and tower shadows Foundational Mix
A vertical pole of length \(6\text{ m}\) casts a shadow \(4\text{ m}\) long. At the same time, a tower casts a shadow \(28\text{ m}\) long. Find the height of the tower.

Step-by-Step Solution

1
Both the pole and tower are vertical (perpendicular to ground), and the sun's rays are parallel (same angle of incidence at the same time). This gives us two right triangles that share the same angle for the sun's rays.
2
By AA similarity, the triangle formed by the pole and its shadow is similar to the triangle formed by the tower and its shadow.
3
Set up the proportion:
\(\dfrac{\text{Height of pole}}{\text{Shadow of pole}} = \dfrac{\text{Height of tower}}{\text{Shadow of tower}}\)
4
Substitute:
\(\dfrac{6}{4} = \dfrac{h}{28}\)
5
Solve:
\(h = \dfrac{6 \times 28}{4} = \dfrac{168}{4} = 42\text{ m}\)
✓ The height of the tower is \(42\text{ m}\).
✦ Verification Check Check the ratios: \(\dfrac{6}{4} = 1.5\) and \(\dfrac{42}{28} = 1.5\). Equal! The similar triangle proportion holds. ✓
P2 Diagonals of a trapezium Basic Mix
Diagonals \(AC\) and \(BD\) of trapezium \(ABCD\) (with \(AB \parallel DC\)) intersect at point \(O\). Using a similarity criterion, show that \(\dfrac{OA}{OC} = \dfrac{OB}{OD}\).

Step-by-Step Solution

1
Since \(AB \parallel DC\), and \(AC\) is a transversal: \(\angle OAB = \angle OCD\) (alternate interior angles).
2
Similarly, with \(BD\) as transversal: \(\angle OBA = \angle ODC\) (alternate interior angles).
3
In \(\triangle AOB\) and \(\triangle COD\): two pairs of angles are equal, so by AA: \(\triangle AOB \sim \triangle COD\).
4
From similar triangles, corresponding sides are proportional:
\(\dfrac{OA}{OC} = \dfrac{OB}{OD} = \dfrac{AB}{CD}\)
✓ Proven: \(\dfrac{OA}{OC} = \dfrac{OB}{OD}\)
P3 Median ratio in similar triangles Intermediate Mix
\(\triangle ABC \sim \triangle PQR\). \(CM\) and \(RN\) are medians of their respective triangles. Prove that \(\dfrac{CM}{RN} = \dfrac{AB}{PQ}\).

Step-by-Step Solution

1
Since \(\triangle ABC \sim \triangle PQR\), we have \(\dfrac{AB}{PQ} = \dfrac{BC}{QR} = \dfrac{CA}{RP}\) and \(\angle A = \angle P\).
2
Since \(CM\) is a median, \(M\) is the midpoint of \(AB\), so \(AM = \dfrac{AB}{2}\). Similarly, \(PN = \dfrac{PQ}{2}\).
3
Therefore: \(\dfrac{AM}{PN} = \dfrac{AB/2}{PQ/2} = \dfrac{AB}{PQ}\). Also, \(\dfrac{CA}{RP} = \dfrac{AB}{PQ}\), so \(\dfrac{AM}{PN} = \dfrac{CA}{RP}\).
4
In \(\triangle AMC\) and \(\triangle PNR\): two pairs of sides proportional with equal included angles (\(\angle A = \angle P\)). By SAS: \(\triangle AMC \sim \triangle PNR\).
5
From this similarity: \(\dfrac{CM}{RN} = \dfrac{CA}{RP} = \dfrac{AB}{PQ}\).
✓ Proven: \(\dfrac{CM}{RN} = \dfrac{AB}{PQ}\)
P4 Isosceles triangle with parallel line Challenging Mix
In isosceles \(\triangle ABC\) with \(AB = AC\), \(D\) is a point on \(BC\) such that \(\angle ADC = \angle BAC\). Show that \(CA^2 = CB \cdot CD\).

Step-by-Step Solution

1
We are given \(\angle ADC = \angle BAC\). Note that \(\angle ACB\) is common to both \(\triangle BAC\) and \(\triangle ADC\) (it's \(\angle ACB = \angle ACD\)).
2
In \(\triangle BAC\) and \(\triangle ADC\): \(\angle BAC = \angle ADC\) (given) and \(\angle ACB = \angle ACD\) (same angle). By AA: \(\triangle BAC \sim \triangle ADC\).
3
From the similarity, corresponding sides are proportional:
\(\dfrac{CA}{CD} = \dfrac{CB}{CA}\)
4
Cross-multiply:
\(CA \times CA = CB \times CD \implies CA^2 = CB \cdot CD\)
✓ Proven: \(CA^2 = CB \cdot CD\)
P5 The grand synthesis — medians and similarity Advanced Mix
If \(AD\) and \(PM\) are medians of \(\triangle ABC\) and \(\triangle PQR\) respectively, where \(\triangle ABC \sim \triangle PQR\), prove that \(\dfrac{AB}{PQ} = \dfrac{AD}{PM}\). (Medians of similar triangles are in the same ratio as corresponding sides.)

Step-by-Step Solution

1
Since \(\triangle ABC \sim \triangle PQR\): \(\dfrac{AB}{PQ} = \dfrac{BC}{QR}\), and \(\angle B = \angle Q\).
2
\(AD\) is a median, so \(D\) is the midpoint of \(BC\): \(BD = \dfrac{BC}{2}\). Similarly, \(QM = \dfrac{QR}{2}\).
3
Now compute the ratio:
\(\dfrac{BD}{QM} = \dfrac{BC/2}{QR/2} = \dfrac{BC}{QR} = \dfrac{AB}{PQ}\)
4
In \(\triangle ABD\) and \(\triangle PQM\): \(\dfrac{AB}{PQ} = \dfrac{BD}{QM}\) (from step 3), and \(\angle B = \angle Q\) (included angles, from similarity). By SAS: \(\triangle ABD \sim \triangle PQM\).
5
From this similarity, all corresponding sides are proportional:
\(\dfrac{AD}{PM} = \dfrac{AB}{PQ}\)
6
This is a beautiful result: medians of similar triangles scale in exactly the same ratio as the corresponding sides. The self-consistent elegance of similarity is complete.
✓ Proven: \(\dfrac{AB}{PQ} = \dfrac{AD}{PM}\)

You have reached the end of Chapter 6 — Triangles.

Similarity is the language of proportion. Every map, photograph, blueprint, and telescope uses this logic.
Trust the ratios. Practice the proofs. The height of no mountain is beyond your reach.