Geometry

high importance~4 Q in Tier 137 formulas⚡ 19 shortcuts6 subtopics

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Congruence, similarity and BPT

⏱ 4 min read🧩 5 question types🎯 12 practice Q

The idea in one minute

Congruent triangles are identical in size and shape. Similar triangles have the same shape but a different size. If the scale factor is $k$ , every length scales by $k$ and every area by $k^2$ . A line parallel to one side cuts the other two sides in the same ratio (BPT). The segment joining two midpoints is half the third side.

Same shape, different size

Similar figures keep the same shape. One is a scaled copy of the other.

If the scale factor is $k$ (a side of the big one is $k$ times the matching side of the small one):

lengths (sides, perimeters, medians, heights) scale by $k$
areas scale by $k^2$

A 3-4-5 triangle scaled by $k=4$ becomes 12-16-20. Its area grows $6 \to 96$ , which is $6\times16=6\times k^2$ .

Rule: Lengths take $k$ once, areas take $k$ twice. That one line is half the subtopic.

From areas back to sides

Going from an area ratio to a length ratio needs a square root.

Areas $25$ and $81$ give a side ratio $5:9$ , never $25:81$ . In the other direction, perimeters $p:q$ give areas $p^2:q^2$ . A $9:25$ area ratio means sides and perimeters in $3:5$ . First match the vertices in order: $A$ pairs with $A'$ , $B$ with $B'$ , so side $AB$ pairs with $A'B'$ . Only then form ratios.

Watch: Options almost always include both the squared and the unsquared ratio. Decide which way you are going before looking.

A line parallel to a side (BPT)

Draw $DE$ parallel to $BC$ inside $\triangle ABC$ . Then

\frac{AD}{DB}=\frac{AE}{EC}

So the parallel line cuts both sides in the same ratio.

If $AD:DB=3:5$ and $AE=6$ cm, then $EC=\dfrac{6\times5}{3}=10$ cm. Cross-multiply and the answer falls out in one line.

Watch: Use part with part ( $AD$ with $DB$ ) or full with full ( $AD$ with $AB$ ). Mixing a part with a full side is the classic slip.

The midpoint theorem

$D$ and $E$ are the midpoints of two sides. Then $DE$ is parallel to the third side and

DE=\frac{BC}{2}

Join all three midpoints and you get four small triangles. Each small side is half a big side, so each small triangle has half the perimeter and one quarter of the area. A 3-4-5 triangle of area $6$ splits into four small triangles of area $1.5$ each. Heights, medians and diagonals are lengths too, so they scale by $k$ just like the sides.

Converse: through the midpoint of one side, a line parallel to a second side bisects the third.

The angle bisector theorem

The bisector of $\angle A$ meets $BC$ at $D$ . Then

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

The opposite side is split in the ratio of its two neighbouring sides.

With $AB=8$ , $AC=12$ , $BC=15$ : the ratio is $8:12=2:3$ , so $DC=\dfrac{3}{5}\times15=9$ cm and $BD=15-9=6$ cm. The two pieces always add back to the whole side, which is a fast check.

Tip: No parallel line is needed here. BPT needs parallels; the bisector theorem needs an angle bisector.

Question types you will see

Each type: how to recognise it, the method step by step, and one question to try.

Type 1very common2 practice Q

Perimeter ratio and area ratio

How to spot it:

Two similar triangles with perimeters (or a pair of matching sides) in a ratio; 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

Method

Find the length ratio from the perimeters or matching sides.
Square it for the area ratio, or take a root to go from areas to lengths.
Scale the known quantity by that ratio.

Why it works:

Area multiplies two lengths, so the scale factor enters twice.

Try this

Two similar triangles have perimeters 24 cm and 36 cm. The larger has area 54 sq cm. Find the smaller area.

Show solution

Side ratio $=\dfrac{24}{36}=\dfrac{2}{3}$ .
Area ratio $=\dfrac{4}{9}$ .
Smaller $=54\times\dfrac{4}{9}=24$ sq cm.

Answer

24 sq cm

Type 2very common2 practice Q

BPT: a parallel line inside the triangle

How to spot it:

$DE\parallel BC$ with three of the four pieces $AD$ , $DB$ , $AE$ , $EC$ known; the fourth is asked.

\frac{AD}{DB}=\frac{AE}{EC}

Method

Write the part ratio on one side equal to the part ratio on the other.
Substitute the three known pieces.
Solve the proportion for the fourth.

Why it works:

The parallel line makes two similar triangles, so both sides carry the same cut ratio.

Try this

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

Show solution

$\dfrac{3}{5}=\dfrac{6}{EC}$ .
$EC=\dfrac{6\times5}{3}$ .
$EC=10$ cm.

Answer

10 cm

Type 3common2 practice Q

Midpoint theorem

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.

DE=\frac{BC}{2}

Method

Confirm both points are midpoints.
Double or halve as asked.
For the triangle of midpoints: perimeter half, area one quarter.

Why it works:

The midpoint segment is the BPT case where the cut ratio is $1:1$ .

Try this

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

Show solution

$XY=\dfrac{QR}{2}$ .
$=\dfrac{14}{2}$ .
$=7$ cm.

Answer

7 cm

Type 4common2 practice Q

Angle bisector theorem

How to spot it:

A bisector meets the opposite side; the two pieces of that side, or a side length, are asked.

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

Method

Write the ratio of the two sides around the bisected angle.
Split the opposite side in that ratio.
With one piece known, scale it by the ratio.

Why it works:

The bisector divides the opposite side exactly as the adjacent sides are divided.

Try this

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

Show solution

$BD:DC=8:12=2:3$ .
$DC=\dfrac{3}{5}$ of $BC$ .
$DC=\dfrac{3}{5}\times15=9$ cm.

Answer

9 cm

Type 5common2 practice Q

Scale a whole triangle

How to spot it:

All three sides of one triangle given; the similar triangle has one side given and the rest asked.

k=\frac{\text{known side of second}}{\text{matching side of first}}

Method

Match corresponding sides: smallest with smallest, largest with largest.
Find the scale factor $k$ once.
Multiply every side or the perimeter by $k$ .

Why it works:

Similarity stretches all lengths by the same factor.

Try this

A triangle with sides 3 cm, 4 cm and 5 cm is similar to a bigger triangle whose hypotenuse is 20 cm. Find the bigger perimeter.

Show solution

$k=\dfrac{20}{5}=4$ .
Sides become $12$ , $16$ , $20$ .
Perimeter $=12+16+20=48$ cm.

Answer

48 cm

Formula sheet

Similarity ratios

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

a = side, P = perimeter, K = area; k = scale factor.

Areas from perimeters

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

Square a length ratio to get the area ratio; take a root to go back.

BPT (Thales)

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

A line parallel to one side cuts the other two sides in the same ratio.

Midpoint theorem

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

The join of two midpoints is half the third side and parallel to it.

Angle bisector theorem

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

The bisector of angle A splits BC in the ratio of the sides AB and AC.

Shortcuts that save time

⚡ Square the perimeter ratio for areas

Perimeters in ratio $p:q$ mean areas in ratio $p^2:q^2$ . Going back, take the square root.

Example

Two similar triangles have perimeters 30 cm and 50 cm. The smaller has area 18 sq cm. Find the larger area.

Show solution

Side ratio $=\dfrac{30}{50}=\dfrac{3}{5}$ .
Area ratio $=\dfrac{9}{25}$ .
Larger $=18\times\dfrac{25}{9}=50$ sq cm.

Answer

50 sq cm

⚡ Spot the word midpoints

'Midpoints of two sides' means the joining segment is half the third side, and parallel to it.

Example

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

Show solution

$BC=2\times DE$ .
$=2\times4.5$ .
$=9$ cm.

Answer

9 cm

⚡ BPT: write the ratio directly

A line parallel to a side cuts equal ratios on both other sides. Write the proportion, substitute, solve.

Example

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

Show solution

$\dfrac{AE}{EC}=\dfrac{AD}{DB}=\dfrac{4}{6}$ .
$EC=\dfrac{5\times6}{4}$ .
$EC=7.5$ cm.

Answer

7.5 cm

Mistakes to avoid

Where most students lose marks on this subtopic.

Mistake 01

Using the side ratio for areas without squaring.

Areas take $k^2$ . A $3:5$ side ratio means a $9:25$ area ratio.

Mistake 02

Squaring when going from areas back to lengths.

That direction needs the square root: areas $25:81$ give sides $5:9$ .

Mistake 03

Using $DE=\dfrac{BC}{2}$ for any parallel line.

The half rule needs midpoints. A general parallel uses BPT ratios.

Mistake 04

Mixing part and full ratios in BPT.

Pair $AD$ with $DB$ , or $AD$ with $AB$ . Never one of each.

Mistake 05

Applying BPT to an angle bisector figure.

No parallel line, no BPT. The bisector splits the side as $AB:AC$ .

Quick revision

Read this the night before the exam.

Similar figures: lengths scale by $k$ , areas by $k^2$ .
Area ratio to length ratio: take the square root.
$DE\parallel BC$ gives $\dfrac{AD}{DB}=\dfrac{AE}{EC}$ .
Midpoints: $DE=\dfrac{BC}{2}$ ; midpoint triangle has half perimeter, quarter area.
Bisector of $\angle A$ : $\dfrac{BD}{DC}=\dfrac{AB}{AC}$ .
Match small with small and large with large before scaling.

Practice: 12 questions

Sets of 10, mixed across the question types above. Every answer has a step-by-step explanation.

Topic test · 12 questions

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

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