Geometry

high importance~4 Q in Tier 136 formulas⚡ 19 shortcuts6 subtopics

Overview Learn (6)Formulas (36)Examples (4)Practice (73)

Subtopic 3 of 6·← Triangles and their centres·Pythagoras theorem and triplets →

Congruence, similarity and BPT

Congruence (same size and shape): SSS, SAS, ASA/AAS, RHS.

Similarity (same shape): corresponding angles equal and sides in the same ratio. For similar triangles with scale factor $k$ :

corresponding sides, perimeters, medians, altitudes → ratio $k$ ,
areas → ratio $k^2$ .

Basic Proportionality Theorem (Thales): a line parallel to one side of a triangle divides the other two sides in the same ratio: $DE\parallel BC\Rightarrow\frac{AD}{DB}=\frac{AE}{EC}$ .

Midpoint theorem: the segment joining midpoints of two sides is parallel to the third side and half its length.

📖 Notes 🧩 5 question types ⚡ Formulas 🎯 12 practice Q

Detailed notes

Congruent vs similar

Congruent triangles are identical in size and shape (SSS, SAS, ASA, RHS). Similar triangles have the same shape: equal corresponding angles and proportional sides. Every question here lives in the proportions, so fix the scale factor $k = \frac{\text{side of first}}{\text{corresponding side of second}}$ and remember what it does:

lengths (sides, perimeters, medians, altitudes) → ratio $k$ ;
areas → ratio $k^2$ .

Going from areas back to lengths needs the square root — the single most-tested switch in this subtopic. Perimeter ratio $p:q$ → area ratio $p^2:q^2$ ; area ratio $a:b$ → perimeter ratio $\sqrt{a}:\sqrt{b}$ . Areas $25$ and $81$ give a side ratio of $5:9$ , never $25:81$ .

Basic Proportionality Theorem (Thales)

Draw a line parallel to one side of a triangle; it cuts the other two sides in the same ratio: $DE \parallel BC \Rightarrow \frac{AD}{DB} = \frac{AE}{EC}$ The full-side form also holds: $\frac{AD}{AB} = \frac{AE}{AC}$ . Keep one form and stay consistent — mixing a part-ratio with a full-ratio is the classic slip. If $AD:DB = 3:4$ , then $E$ also cuts $AC$ as $3:4$ , so $AE = \frac{3}{7}AC$ .

Midpoint theorem

If $D, E$ are the midpoints of two sides, $DE$ is parallel to the third side and $DE = \frac{BC}{2}$ . The triangle formed by joining all three midpoints has sides exactly half of the original — its perimeter is half and its area is one-quarter of the original. Converse: through the midpoint of one side, a line parallel to a second side bisects the third.

Angle bisector theorem

The internal bisector of $\angle A$ meets $BC$ at $D$ with $\frac{BD}{DC} = \frac{AB}{AC}$ This divides the opposite side in the ratio of the adjacent sides — a different mechanism from BPT (no parallel line involved), though the number work looks identical. With $BC$ known, split it in the ratio; with one part known, scale the other.

Reading similar-triangle data

When one full set of sides is known and one side of the twin: match the corresponding sides (largest with largest, smallest with smallest), get $k$ once, then scale everything by $k$ . When perimeters and areas are mixed, convert perimeters to a side ratio, square it for areas, or take the root to go back.

Quick revision

Similar: lengths $\times k$ ; areas $\times k^2$ . Areas → lengths: take $\sqrt{\ }$ .
$DE \parallel BC \Rightarrow \frac{AD}{DB} = \frac{AE}{EC}$ .
Midpoints: $DE = \frac{BC}{2}$ ; midpoint triangle: perimeter $\frac12$ , area $\frac14$ .
Bisector of $A$ : $\frac{BD}{DC} = \frac{AB}{AC}$ .
Match corresponding sides (small ↔ small, large ↔ large) before scaling.

Types of questions asked

Every way this subtopic shows up in exams — how to recognise it, the formula or logic to use, and a solved example.

Type 1: Perimeter ratio ↔ area ratiovery common2 practice Q

How to spot it:

Two similar triangles have perimeters (or a pair of corresponding sides/altitudes) in a given ratio and one area is asked — or areas are given and a perimeter asked.

\frac{K_1}{K_2} = \left(\frac{P_1}{P_2}\right)^2

Perimeters (sides, altitudes) give the linear ratio $k$ .
Areas need $k^2$ ; recovering lengths from areas needs $\sqrt{k_{\text{area}}}$ .
Scale the known area/perimeter by the right ratio.

Why: area is two lengths multiplied, so the scale factor enters twice.

Example: Two similar triangles have perimeters 24 cm and 36 cm. If the area of the larger triangle is 54 sq cm, find the area of the smaller.

Side ratio $= \frac{24}{36} = \frac{2}{3}$ , so area ratio $= \frac{4}{9}$ : smaller area $= 54 \times \frac{4}{9} = 24$ sq cm.

Type 2: BPT: a parallel line inside the trianglevery common2 practice Q

How to spot it:

$DE \parallel BC$ is stated (or drawn) with three of the four segments $AD, DB, AE, EC$ given; the fourth is asked.

DE \parallel BC \Rightarrow \frac{AD}{DB} = \frac{AE}{EC}

Write the part-ratio on one side equals the part-ratio on the other.
Substitute the three known segments; solve the proportion.
If a full side is given instead, switch to $\frac{AD}{AB} = \frac{AE}{AC}$ — but never mix part with full.

Why: the parallel line makes two similar triangles, so the cut points keep one ratio on both sides.

Example: In $\triangle ABC$ , $DE \parallel BC$ with $AD = 3$ cm, $DB = 5$ cm and $AE = 6$ cm. Find $EC$ .

$\frac{AD}{DB} = \frac{AE}{EC} \Rightarrow \frac{3}{5} = \frac{6}{EC} \Rightarrow EC = 10$ cm.

Type 3: Midpoint theoremcommon2 practice Q

How to spot it:

'D and E are the midpoints of AB and AC' — find $DE$ from $BC$ or $BC$ from $DE$ , or use the midpoint-triangle perimeter/area.

DE = \frac{BC}{2}

Midpoints on two sides → the joining segment is half the third side and parallel to it.
Double/halve as asked.
All three midpoints joined: each small side is half a big side → perimeter $\frac12$ , area $\frac14$ .

Why: the midpoint segment is the special BPT case with ratio $1:1$ .

Example: In $\triangle PQR$ , $X$ and $Y$ are the midpoints of $PQ$ and $PR$ . If $QR = 14$ cm, find $XY$ .

$XY = \frac{QR}{2} = 7$ cm by the midpoint theorem.

Type 4: Angle bisector theoremcommon2 practice Q

How to spot it:

The bisector of an angle of a triangle meets the opposite side; the two parts of that side (or a side length) are asked.

\frac{BD}{DC} = \frac{AB}{AC}

The bisected angle's two sides give the ratio $AB : AC$ .
The opposite side is split in that same ratio.
With $BC$ known, divide it in the ratio; with one part known, scale.

Why: the bisector creates two triangles that mirror the ratio via the sine rule — the exam-ready statement is the ratio formula.

Example: In $\triangle ABC$ , the bisector of $\angle A$ meets $BC$ at $D$ . If $AB = 8$ cm, $AC = 12$ cm and $BC = 15$ cm, find $DC$ .

$BD:DC = 8:12 = 2:3$ ; $DC = \frac{3}{5} \times 15 = 9$ cm.

Type 5: Scaling a full triangle (find the missing sides)common2 practice Q

How to spot it:

One triangle's three sides are given; a similar triangle has one side known — the remaining sides or its perimeter is asked.

k = \frac{\text{known side of 2nd}}{\text{corresponding side of 1st}}

Match corresponding sides: smallest ↔ smallest, largest ↔ largest.
Compute the scale $k$ once.
Multiply every side (or the whole perimeter) by $k$ .

Why: similarity scales all lengths by the same $k$ — one multiplication per side, no per-side algebra.

Example: A triangle with sides 3, 4 and 5 cm is similar to a larger triangle whose hypotenuse is 20 cm. The perimeter of the larger triangle is:

$k = \frac{20}{5} = 4$ ; sides $12, 16, 20$ → perimeter $= 4 \times 12 = 48$ cm.

Formulas

Similarity ratios

\frac{a_1}{a_2}=k,\quad \frac{P_1}{P_2}=k,\quad \frac{K_1}{K_2}=k^2

Areas from perimeters

\frac{K_1}{K_2}=\left(\frac{P_1}{P_2}\right)^2

BPT

DE\parallel BC\Rightarrow\frac{AD}{DB}=\frac{AE}{EC}

Midpoint theorem

D,E\ \text{midpoints}\Rightarrow DE=\frac{BC}{2}

Angle bisector theorem

\frac{BD}{DC}=\frac{AB}{AC}\ \text{(internal bisector of }A\text{)}

Shortcut tricks

⚡ Square the perimeter ratio for areas

Perimeters in ratio $p:q$ → areas in ratio $p^2:q^2$ (and back: side ratio $=\sqrt{\text{area ratio}}$ ).

Example: Two similar triangles have perimeters 30 cm and 50 cm. If the area of the smaller is 18 sq cm, find the area of the larger.

Side ratio $=\frac35$ , so area ratio $=\frac{9}{25}$ : larger area $=18\times\frac{25}{9}=50$ sq cm.

⚡ Midpoint theorem shortcut

Spot the words "midpoints of two sides" → the joining segment is half the third side, parallel to it. Conversely, through the midpoint of one side, parallel to another → it bisects the third side.

Example: In $\triangle ABC$ , $D$ and $E$ are midpoints of $AB$ and $AC$ . If $DE=4.5$ cm, find $BC$ .

$BC=2\times DE=9$ cm.

⚡ BPT with a parallel drawn inside

When a line parallel to a side cuts the other two sides, write the ratio directly. If the segments on one side are $AD:DB=2:3$ , the same ratio holds on the other side.

Example: In $\triangle ABC$ , $DE\parallel BC$ with $AD=4$ cm, $DB=6$ cm, $AE=5$ cm. Find $EC$ .

$\frac{AE}{EC}=\frac{AD}{DB}=\frac46\Rightarrow EC=\frac{5\times6}{4}=7.5$ cm.

Where students lose marks

Using the side ratio (not its square) for areas — or squaring when you should not.
Taking $DE=\frac{BC}{2}$ as true for any parallel line; it needs midpoints.
BPT ratio written as $AD/AB=AE/AC$ carelessly — it is correct, but mixing $AD/DB$ with $AE/AC$ on opposite sides is the classic slip.

Practice sets — 12 questions

Sets of 10, mixed across the question types above. Each answer comes with a step-by-step explanation.

Topic test · 12 questions

Suggested time 8 min · wrong answers go to your mistake notebook automatically.