Circle Geometry, Tangents, Trapezium & Polygon

Circle geometry covers key theorems involving chords, tangents, triangles with medians, trapeziums, and polygons. These results are frequently tested in SSC, Banking, and Railway exams.

Triangle: Midpoint / Median Relations

If D is the midpoint of BC and E is the midpoint of AB in triangle ABC, then the following relations hold

In triangle ABC, E is midpoint of AB and D is midpoint of BC. If AC = 10, find $ED = \frac12 AC = \frac 12 10 = 5$

D and E must be midpoints of their respective sides.
ED is always half of AC (Midpoint Theorem).

Perpendicular Medians

When Two Medians Are Perpendicular If medians BE and CD of triangle ABC are perpendicular to each other,

then: $AB^2 + AC^2 = 5BC^2$

if BE ⊥ CD (two medians are perpendicular), the formula always becomes: > AB² + AC² = 5BC² > The side between the two median feet (BC) gets multiplied by 5.

Direct & Transverse Common Tangents

3. Direct and Transverse Common Tangents For two circles with radii r₁ and r₂, and distance between centres = d, and tangent length = ℓ:

Direct Common Tangent: $ℓ^2 = d^2 − (r₁ − r₂)^2$

Transverse Common Tangent: $ℓ^2 = d^2 − (r₁ + r₂)^2$

• Direct tangent: does NOT cross between the circles.
• Transverse tangent: CROSSES between the two circles.
• If circles touch externally, number of common tangents = 3.
• If circles touch internally, number of common tangents = 1.

4. Trapezium — Midpoint Segment

In trapezium ABCD where AB ∥ DC, if P and Q are the midpoints of the non-parallel sides AD and BC respectively, then:

$PQ = (AB − DC) / 2$

In trapezium ABCD, AB = 14 cm and DC = 6 cm. Find PQ (segment joining midpoints of non-parallel sides).

PQ = (AB − DC) / 2 = (14 − 6) / 2 = 4 cm

Polygon: Number of Diagonals

For any polygon with n sides:

Number of Diagonals = $\frac{n(n-3)}{2}$

For a hexagon (n = 6):

Diagonals = $\frac{6(6-3)}{2} = \frac{6 \times 3}{2} = 9$

Polygon | n | Diagonals

Triangle | 3 | 0 , Quadrilateral | 4 | 2 , Pentagon | 5 | 5 , Hexagon | 6 | 9 , Octagon | 8 | 20

Circle Angle: Interior & Exterior Point

Case 1 — Point P inside the circle (chords intersect inside): $\angle BPD = \frac{\text{arc } x + \text{arc } y}{2}$

Case 2 — Point P outside the circle (secants from outside): $\angle BPD = \frac{\overset{\frown}{x} – \overset{\frown}{y}}{2}$

Key Points:

• arc x = the larger intercepted arc, arc y = the smaller.
• Inside the circle → ADD the two arcs, divide by 2.
• Outside the circle → SUBTRACT the two arcs, divide by 2.
• Frequently tested in SSC CGL Tier 1 and Tier 2.