SSC CGL Maths · Triangles

Parallels, Midpoints and the Bisector Theorem

Draw a line parallel to a side and the other two sides are divided proportionally; join two midpoints and you get half the third side; bisect an angle and the far side divides as the two arms of that angle do.

All Triangles notes

Why this matters

Thirteen questions in the bank, all EASY. Each is one ratio set up correctly: four use the parallel-line ratio, six the midpoint theorem and the quarter-area triangle it makes, and three the angle bisector theorem. The wrong options are the answers you get from the right numbers in the wrong ratio.

Concept 1 of 3: Basic proportionality theorem: a line parallel to one side

A line drawn across a triangle parallel to its base cuts off a smaller similar triangle on top. Because the top triangle is a scaled copy, the two sides are cut in the same proportion.

Definition

DE ∥ BC with D on AB and E on AC: $\frac{AD}{DB} = \frac{AE}{EC}$ . This is the basic proportionality theorem (Thales).
Also $\frac{AD}{AB} = \frac{AE}{AC} = \frac{DE}{BC}$ , from the similar triangles ADE and ABC. The parallel segment DE goes with the whole side AB, not with DB.
Converse: if $\frac{AD}{DB} = \frac{AE}{EC}$ , then DE is parallel to BC.
A condition like $5AE = 3EC$ means AE : EC = 3 : 5: the larger number goes with the smaller piece.

Basic proportionality theorem

\frac{AD}{DB} = \frac{AE}{EC} \qquad \frac{DE}{BC} = \frac{AD}{AB}

D, Epoints on AB and AC with DE parallel to BC

Worked example

In triangle PQR, ST ∥ QR with S on PQ and T on PR. PS = 4 cm, SQ = 10 cm and PR = 21 cm. Find PT and TR.

Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

SSC CGL · 2025 · Tier I, 26 Sep 2025, Shift 1 · Q71Easy

Example 1 · Triangles · Parallels, Midpoints and the Bisector Theorem

In triangle ABC, DE is drawn parallel to BC, intersecting AB and AC at D and E respectively. If AD=3, DB=6, and AE=4, find EC.

Using AD : DB for the parallel segment

DE and BC are sides of the similar triangles ADE and ABC, so DE : BC = AD : AB. With AD : DB = 1 : 2, DE is a third of BC, not a half.

Reading 5AE = 3EC as AE : EC = 5 : 3

If five of AE equal three of EC, then AE is the shorter piece: AE : EC = 3 : 5.

Drill 3 more on basic proportionality theorem: a line parallel to one side

Concept 2 of 3: The midpoint theorem and the quarter triangle

Joining the midpoints of two sides is the parallel-line case with the ratio 1 : 1. The cut-off triangle is a half-size copy, so the joining segment is half the third side.

Definition

Midpoint theorem: the segment joining the midpoints of two sides is parallel to the third side and half as long.
Converse: a line through the midpoint of one side, parallel to a second side, passes through the midpoint of the third.
Joining all three midpoints cuts the triangle into four congruent triangles (each pair shares sides by the midpoint theorem, so SSS). The middle one is not congruent to the whole; it is similar with $k = \frac{1}{2}$ .
So each small triangle has $\frac{1}{4}$ of the area and half the perimeter of the whole. The quadrilateral left after cutting off one corner triangle has $\frac{3}{4}$ of the area.
The midpoint triangle of an equilateral triangle is equilateral, with half the side.

Midpoint theorem

XY = \frac{1}{2}BC \qquad \text{ar}(XYZ) = \frac{1}{4}\,\text{ar}(ABC)

X, Ythe midpoints of AB and AC
XYZthe triangle joining the midpoints of all three sides

Worked example

In triangle ABC, X and Y are the midpoints of AB and AC, BC = 18 cm and the area of ABC is 60 cm². Find XY, the area of triangle AXY and the area of the quadrilateral XYCB.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

SSC CGL · 2024 · Tier I, 11 Sep 2024, Shift 2 · Q55Easy

Example 2 · Triangles · Parallels, Midpoints and the Bisector Theorem

\triangle

PQR, M, N, and S are the mid-points of sides PQ, PR and QR, respectively. If the area of the

\triangle

PQR is 46

\text{cm}^2

, what is the area (in

\text{cm}^2

) of the quadrilateral MNRQ?

Taking the midpoint triangle as half the area

The midpoint triangle has half the side lengths, so it has a quarter of the area. Half is the ratio of perimeters.

Calling the midpoint triangle congruent to the whole

The midpoint triangle is congruent to the three corner triangles beside it, and only similar to the original triangle.

Drill 5 more on the midpoint theorem and the quarter triangle

Concept 3 of 3: The angle bisector theorem

The bisector of an angle leans towards the shorter side next to it, so it meets the opposite side closer to that end. The split matches the two sides exactly.

Definition

The bisector of $\angle A$ meets BC at D. Then $\frac{BD}{DC} = \frac{AB}{AC}$ : each piece goes with the side that touches it.
So $BD = \frac{AB}{AB + AC} \times BC$ and $DC = \frac{AC}{AB + AC} \times BC$ .
The length of BC is only used at the end, to turn the ratio into lengths.
If AB = AC the bisector meets BC at its midpoint.

Angle bisector theorem

\frac{BD}{DC} = \frac{AB}{AC} \qquad BD = \frac{AB}{AB + AC} \times BC

Dthe point where the bisector of angle A meets BC

Worked example

In triangle PQR, PQ = 15 cm, PR = 10 cm and QR = 20 cm. The bisector of

\angle P

meets QR at S. Find QS and SR.

Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

SSC CGL · 2024 · Tier I, 12 Sep 2024, Shift 2 · Q54Easy

Example 3 · Triangles · Parallels, Midpoints and the Bisector Theorem

\Delta DEF

, the bisector of

\angle D

intersects side EF at point N. If DE = 36 cm, DF = 40 cm and EF = 38 cm, then the length (in cm) of NF is ________.

Pairing each piece with the far side

BD touches B, so it goes with AB; DC goes with AC. Swapping them gives the right numbers in the wrong order, and that answer is usually printed.

Putting the third side into the ratio

The ratio uses only the two sides that meet at the bisected angle. The side being cut is used afterwards, to share it out.

Drill 2 more on the angle bisector theorem

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Basic proportionality theorem: a line parallel to one side
Basic proportionality theorem
$\frac{AD}{DB} = \frac{AE}{EC} \qquad \frac{DE}{BC} = \frac{AD}{AB}$
The midpoint theorem and the quarter triangle
Midpoint theorem
$XY = \frac{1}{2}BC \qquad \text{ar}(XYZ) = \frac{1}{4}\,\text{ar}(ABC)$
The angle bisector theorem
Angle bisector theorem
$\frac{BD}{DC} = \frac{AB}{AC} \qquad BD = \frac{AB}{AB + AC} \times BC$

Watch out for (6)

Using AD : DB for the parallel segment
→ Basic proportionality theorem: a line parallel to one side
Reading 5AE = 3EC as AE : EC = 5 : 3
→ Basic proportionality theorem: a line parallel to one side
Taking the midpoint triangle as half the area
→ The midpoint theorem and the quarter triangle
Calling the midpoint triangle congruent to the whole
→ The midpoint theorem and the quarter triangle
Pairing each piece with the far side
→ The angle bisector theorem
Putting the third side into the ratio
→ The angle bisector theorem

Test yourself on Triangles

20 past SSC CGL questions from this chapter, timed at 12 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

Take the Triangles chapter test Drill all 13 questions