SSC CGL Preparation: Concepts and Practice

Learning outcome

Distinguish medians, altitudes and angle bisectors, and locate four triangle centres.

Concepts and assumptions

Use a nondegenerate triangle ABC. A median joins a vertex to the opposite side’s midpoint. It creates equal-area triangles because their bases are equal and perpendicular heights match. A median need not be perpendicular or bisect an angle.

Medians meet at centroid G, always inside. On median AM, AG:GM = 2:1; AG is two-thirds of AM. Comparing equal-area subdivisions made by the medians gives this ratio.

An altitude is perpendicular from a vertex to the opposite side’s line, possibly reaching an extension. Altitude lines meet at orthocentre H: inside an acute triangle, at the right-angle vertex in a right triangle, and outside an obtuse triangle.

Internal angle bisectors halve vertex angles and meet at incentre I, always inside. A bisector’s points have equal perpendicular distances from its arms. Thus I is equally distant from all three side lines and is the inscribed circle’s centre.

If AD bisects ∠A and D lies on BC, the angle-bisector theorem gives BD/DC = AB/AC: adjacent sides determine the opposite side’s division.

A side’s perpendicular bisector passes through its midpoint at 90°. Its points are equidistant from that side’s endpoints. These bisectors meet at circumcentre O, where OA = OB = OC. O is inside an acute triangle, at the hypotenuse midpoint in a right triangle, and outside an obtuse triangle.

All four centres coincide in an equilateral triangle, but not generally.

Worked examples

Example 1 — Divide a median. AM is a median of triangle ABC, M lies on BC, and AM = 18 cm. For centroid G, one ratio part = 18/3 = 6 cm. Hence AG = 12 cm and GM = 6 cm.

Example 2 — Right-triangle centres. Triangle PQR is right-angled at Q with hypotenuse PR = 18 cm. Circumcentre O is PR’s midpoint, so OP = OR = OQ = 9 cm. Orthocentre H is Q. Centroid and incentre remain strictly inside.

Example 3 — Divide a side. In triangle ABC, AB = 10 cm, AC = 15 cm and BC = 20 cm. Internal bisector AD meets BC at D. Then BD:DC = 10:15 = 2:3. One part = 20/5 = 4 cm, giving BD = 8 cm and DC = 12 cm.

Common mistakes

Perpendicular bisectors and altitudes have different definitions, even when they coincide. Distances from sides are perpendicular distances. Not every centre is inside, and a median need not bisect an angle.

Practice questions

1. G is the centroid on median AM, and AG = 14 cm. Find GM and AM.
2. Incentre I is 4 cm perpendicularly from side AB. Find its distances from AC and BC and the inscribed-circle radius.
3. In triangle ABC, AB = 9 cm, AC = 12 cm and BC = 14 cm. Internal bisector AD meets BC at D. Find BD and DC.
4. Triangle PQR has ∠P = 108° and ∠Q = 42°. Locate its centroid, incentre, circumcentre and orthocentre: inside or outside?

Worked solutions

1. AG represents two parts, so GM = 14/2 = 7 cm. AM = 14 + 7 = 21 cm.
2. Equal perpendicular distances give 4 cm for both AC and BC. The inscribed-circle radius is also 4 cm.
3. BD:DC = 9:12 = 3:4. One part = 14/7 = 2 cm, so BD = 6 cm and DC = 8 cm.
4. The third angle is 180° - 108° - 42° = 30°. The triangle is obtuse: centroid and incentre are inside; circumcentre and orthocentre are outside.