MathsICSE Class 9 · Area Theorems

Medians, Equal Areas and Equal Altitudes: Splitting Triangles by Area (ICSE Class 9)

Updated 17 October 2026 · Fact-checked

A median divides a triangle into two triangles of equal area. Triangles with equal areas on the same base (or equal bases) have equal altitudes. When two triangles share the same height, the ratio of their areas equals the ratio of their bases, which lets you split a triangle in any given ratio.

Median AD splits the base into equal parts BD = DC, and both triangles share the same height h, so each has half the area of ABC.

Illustration drawn for QuestionBuddy.

Understand Medians, Equal Areas and Equal Altitudes of Triangles

This topic uses the area formula $ar = \frac{1}{2} \times base \times height$ in the other direction: instead of finding the area, you read off what equal areas tell you about heights and bases.

Result A (median). A median $A D$ of $△ A B C$ divides it into $△ A B D$ and $△ A C D$ with equal bases $B D = D C$ and the same height from $A$ . So
$ $ar (△ A B D) = ar (△ A C D) = \frac{1}{2} ar (△ A B C) .$ $

Result B (equal altitudes). If two triangles stand on the same base and have equal areas, their corresponding altitudes are equal, because $\frac{1}{2} \times b \times h_{1} = \frac{1}{2} \times b \times h_{2}$ gives $h_{1} = h_{2}$ . If they are on the same side of the base, the apexes lie on a line parallel to the base.

Result C (ratio of areas). Triangles with the same height have areas in the ratio of their bases. If $D$ is on $B C$ with $B D : D C = m : n$ , then
$ $ar (△ A B D) : ar (△ A D C) = m : n .$ $
This is the tool for splitting a triangle in a given ratio and for finding unknown parts.

Using medians repeatedly. If $D$ is the mid-point of $B C$ and $E$ is the mid-point of $A D$ , then $B E$ is a median of $△ A B D$ and $C E$ is a median of $△ A C D$ , so $ar (△ B E D) = \frac{1}{4} ar (△ A B C)$ and $ar (△ B E C) = \frac{1}{2} ar (△ A B C)$ .

Point on a median. If $P$ is any point on median $A D$ , then $P D$ is a median of $△ P B C$ , so $ar (△ P B D) = ar (△ P C D)$ . Subtracting from the equal triangles $A B D$ and $A C D$ gives $ar (△ A B P) = ar (△ A C P)$ .

Writing it up. State the reason beside each equality: " $A D$ is a median", "same height", "equal bases". Do not just write numbers; the reason carries the marks.

Medians, Equal Areas and Equal Altitudes of Triangles: formulas to remember

Median halves the triangle: $ar (△ A B D) = ar (△ A C D) = \frac{1}{2} ar (△ A B C)$; D is the mid-point of BC.
Equal areas, same base: $ar (△ A B C) = ar (△ P B C) \Rightarrow h_{A} = h_{P}$; Equal altitudes on the common base.
Same height, ratio of bases: $\frac{ar ( △ A B D )}{ar ( △ A D C )} = \frac{B D}{D C}$; D on BC.
Point on a median: $ar (△ A B P) = ar (△ A C P)$; P on median AD.

How to solve Medians, Equal Areas and Equal Altitudes of Triangles questions

Turn every statement into "same height" or "equal bases", then compare areas.

1Mark the common height or the equal bases on the figure.
2If a median or mid-point appears, write the equal-area pair it creates.
3If the ratio of segments on a side is given, use it as the ratio of the areas of the two triangles with the opposite vertex.
4Express every unknown region in terms of the known total or parts.
5Add the parts back to check the total.

Medians, Equal Areas and Equal Altitudes of Triangles in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Splitting a plot fairly

Suppose two brothers inherit a triangular plot $A B C$ of area 400 m². Their rule is to draw a line from vertex $A$ to a point $D$ on the opposite side $B C$ .

Where must $D$ be so that both brothers get equal land? If instead $B D : D C = 3 : 2$ , how much land does each get?

Try it first, then show the solutionSolution

Equal land needs equal areas with the same height from $A$ , so $B D = D C$ : $D$ is the mid-point of $B C$ and $A D$ is a median.
With $B D : D C = 3 : 2$ , areas are in the ratio $3 : 2$ : $\frac{3}{5} \times 400 = 240$ m² and $\frac{2}{5} \times 400 = 160$ m².

Answer: Mid-point of $B C$ for equal shares; 240 m² and 160 m² for the 3 : 2 split.

Memory hook: Same height means area splits like the base.

Sail cloth cut

Imagine a triangular sail of area 30 square units cut from a base of 12 units; a designer wants another triangular sail on the same base with the same area.

How high above the base must the new apex be, and where can it lie?

Try it first, then show the solutionSolution

$\frac{1}{2} \times 12 \times h = 30$ , so $h = 5$ .
The apex can be anywhere on the line parallel to the base at distance 5 on the same side.

Answer: Height 5 units; apex anywhere on the parallel line at that distance.

Memory hook: Equal area on a common base means equal altitude.

Tricks to remember Medians, Equal Areas and Equal Altitudes of Triangles

Short memory aids. Say each one aloud once, then use it on a question the same day.

Same height, share the base
Two triangles under the same apex on one line have areas in the ratio of their bases.
Use it when: Splitting a triangle by a cevian.
Median = halving
A median always halves the area, whatever the shape of the triangle.
Use it when: Starting any median question.
Halve again
Mid-point of a median creates quarter areas on each half.
Use it when: Chains of mid-points.

Medians, Equal Areas and Equal Altitudes of Triangles: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Median and area · 1 mark · MCQBoard level

$A D$ is a median of $△ A B C$ and $ar (△ A B D) = 37$ cm². Then $ar (△ A B C)$ is:
(a) 18.5 cm²
(b) 37 cm²
(c) 74 cm²
(d) 111 cm²

Try it first, then show the solutionSolution

A median halves the triangle: $ar (△ A B C) = 2 \times 37 = 74$ cm².

Answer: (c) 74 cm²

Example 2: Equal altitudes · 3 marks · Short answerBoard level

Triangles $A B C$ and $D B C$ are on the same base $B C = 10$ cm and each has area $40$ cm². Find the altitude of each triangle on $B C$ . Where does $D$ lie if it is on the same side of $B C$ as $A$ ?

Try it first, then show the solutionSolution

$\frac{1}{2} \times 10 \times h = 40$ , so $h = 8$ cm for both.
Equal altitudes on the same side mean $D$ is at distance 8 cm from $B C$ , so $D$ lies on the line through $A$ parallel to $B C$ .

Answer: Altitude $8$ cm each; $A D ∥ B C$ .

Example 3: Mid-point of a median · 3 marks · Short answerHarder board level

In $△ A B C$ with $ar (△ A B C) = 96$ cm², $D$ is the mid-point of $B C$ and $E$ is the mid-point of $A D$ . Find $ar (△ B E D)$ and $ar (△ B E C)$ .

Try it first, then show the solutionSolution

$A D$ is a median: $ar (△ A B D) = 48$ cm².
$B E$ is a median of $△ A B D$ : $ar (△ B E D) = 24$ cm².
$C E$ is a median of $△ A C D$ so $ar (△ C E D) = 24$ cm²; $ar (△ B E C) = 24 + 24 = 48$ cm² $= \frac{1}{2} ar (△ A B C)$ .

Answer: $ar (△ B E D) = 24$ cm²; $ar (△ B E C) = 48$ cm².

Example 4: Dividing in a ratio · 4 marks · Short answerHarder board level

In $△ A B C$ of area $100$ cm², $P$ lies on $A B$ with $A P : P B = 3 : 2$ and $Q$ is the mid-point of $P C$ . Find $ar (△ A P C)$ , $ar (△ P B C)$ and $ar (△ A P Q)$ .

Try it first, then show the solutionSolution

$△ A P C$ and $△ P B C$ have the same height from $C$ to $A B$ , so their areas are in the ratio $A P : P B = 3 : 2$ .
$ar (△ A P C) = \frac{3}{5} \times 100 = 60$ cm²; $ar (△ P B C) = \frac{2}{5} \times 100 = 40$ cm².
$A Q$ is a median of $△ A P C$ ( $Q$ is the mid-point of $P C$ ): $ar (△ A P Q) = \frac{1}{2} \times 60 = 30$ cm².

Answer: $60$ cm², $40$ cm², $30$ cm².

Example 5: Any point on a median · 4 marks · Long answerHOTS / competency

$P$ is any point on the median $A D$ of $△ A B C$ . Prove that $ar (△ A B P) = ar (△ A C P)$ . If $ar (△ A B C) = 80$ cm² and $ar (△ P B C) = 30$ cm², find $ar (△ A B P)$ .

Try it first, then show the solutionSolution

$A D$ is a median of $△ A B C$ : $ar (△ A B D) = ar (△ A C D)$ ... (1).
$P D$ is a median of $△ P B C$ : $ar (△ P B D) = ar (△ P C D)$ ... (2).
Subtract (2) from (1): $ar (△ A B P) = ar (△ A C P)$ . Hence proved.
The two equal triangles together are $80 - 30 = 50$ cm², so $ar (△ A B P) = 25$ cm².

Answer: Proved; $ar (△ A B P) = 25$ cm².

Where students lose marks in Medians, Equal Areas and Equal Altitudes of Triangles

The slips examiners see most often, and what to write instead.

Mistake: Using the ratio of sides as the ratio of areas without a common height
The area ratio equals the base ratio only if the heights are the same.
Wrong working
$B D : D C = 3 : 5$ , so $ar (△ A B D) : ar (△ A B C) = 3 : 5$ .
Right working
$ar (△ A B D) : ar (△ A D C) = 3 : 5$ , and $ar (△ A B D) : ar (△ A B C) = 3 : 8$ .
Do this: Compare the two parts, then divide by the total.
Usually costs: Final answer.
Mistake: Calling any line from a vertex a median
Only the line to the mid-point halves the area.
Wrong working
$A D$ divides $B C$ in ratio $1 : 3$ , so $ar (△ A B D) = \frac{1}{2} ar (△ A B C)$ .
Right working
$ar (△ A B D) = \frac{1}{4} ar (△ A B C)$ for a $1 : 3$ split.
Do this: Check BD = DC first.
Usually costs: Two or three marks.
Mistake: Stopping without the reason
Each equality needs its reason in words.
Wrong working
$ar (△ A B P) = ar (△ A C P)$ .
Right working
$ar (△ A B P) = ar (△ A C P)$ because $A D$ and $P D$ are medians of $△ A B C$ and $△ P B C$ , and subtracting equal areas.
Do this: Write the reason beside each step.
Usually costs: Half the marks of a proof.

Exam tips

Class IX is examined in school papers; the harder items here are the sort used for foundation-olympiad practice and are labelled as such.
In ratio questions, write the ratio as a fraction of the total (3 and 5 give 3/8 and 5/8) before multiplying.
A figure with two or three mid-points is almost always a repeated-median question: halve step by step.
Check your final answer by adding the pieces: they must sum to the whole triangle.

Practice questions on Medians, Equal Areas and Equal Altitudes of Triangles

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
Triangles $A B C$ and $D B C$ lie on the same side of $B C$ and have equal areas. Then:
(a) $A D ∥ B C$
(b) $A D ⊥ B C$
(c) $A B = D B$
(d) $A C = D C$
Hint
Equal area on the same base gives equal altitudes.
Answer
(a) AD is parallel to BC.
Q2 · 3 marks · Short answerHarder board level
In $△ A B C$ of area $120$ cm², $D$ and $E$ are the mid-points of $B C$ and $A C$ . Find $ar (△ C D E)$ and $ar (A B D E)$ .
Hint
$△ A D C$ is half; $D E$ is a median of it.
Answer
$ar (△ A D C) = 60$ cm²; $D E$ is a median of $△ A D C$ so $ar (△ C D E) = 30$ cm². $ar (A B D E) = 120 - 30 = 90$ cm².
Q3 · 3 marks · Short answerHarder board level
$Q$ is a point on $B C$ of $△ A B C$ with $B Q = 2$ cm, $QC = 5$ cm and $ar (△ A B Q) = 18$ cm². Find $ar (△ A QC)$ and $ar (△ A B C)$ .
Hint
Same height from A: area ratio = base ratio 2 : 5.
Answer
$\frac{18}{ar ( △ A QC )} = \frac{2}{5}$ , so $ar (△ A QC) = 45$ cm² and $ar (△ A B C) = 63$ cm².
Q4 · 1 mark · Assertion-reasonHarder board level
Assertion (A): A median of a triangle divides it into two triangles of equal area.
Reason (R): The two triangles have equal bases and the same altitude from the vertex.
(a) Both (A) and (R) are true and (R) explains (A).
(b) Both are true but (R) does not explain (A).
(c) (A) is true, (R) is false.
(d) (A) is false, (R) is true.
Hint
Equal bases, same height.
Answer
(a)
Q5 · 4 marks · Case-basedHOTS / competency
Case study: A triangular park $A B C$ has area $480$ m². A path from $A$ meets $B C$ at $D$ with $B D : D C = 3 : 5$ , and a bench $E$ is at the mid-point of the path $A D$ .
(a) Find $ar (△ A B D)$ .
(b) Find $ar (△ A E C)$ .
(c) Find $ar (△ B E C)$ and compare it with half the park.
Hint
Ratio of bases gives ABD and ADC; E halves each.
Answer
(a) $\frac{3}{8} \times 480 = 180$ m² (b) $ar (△ A D C) = 300$ , $C E$ is a median of $△ A D C$ so $ar (△ A E C) = 150$ m² (c) $ar (△ B E D) = 90$ , $ar (△ C E D) = 150$ , so $ar (△ B E C) = 240$ m² $= \frac{1}{2} \times 480$ .

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.What does a median do to the area of a triangle?Show answer
It divides it into two triangles of equal area.
2.Triangles with the same height: how do their areas compare?Show answer
In the ratio of their bases.
3.What follows if two triangles on the same base have equal area?Show answer
Their altitudes on that base are equal.
4.P lies on median AD. What can you say about ar(ABP) and ar(ACP)?Show answer
They are equal.

Medians, Equal Areas and Equal Altitudes of Triangles: frequently asked questions

Is the equal-altitude result part of the ICSE Class IX syllabus?

Yes. The syllabus lists that triangles of equal area on the same bases have equal corresponding altitudes.

Can I use the ratio of areas result without proof?

Yes, it follows directly from the area formula; state "same height" as your reason.

Why does the mid-point of a median give quarter areas?

Because each half of the triangle is halved again by the median of that half.