Division of line segments, construction of tangents to a circle
In Class 10 Constructions, we use only compass and ruler (straightedge). We learn two major constructions: dividing a line segment in a given ratio, and drawing tangents to a circle from an external point.
Geometric constructions are the basis of engineering drawings, architecture blueprints, and CAD design. Understanding how to precisely divide segments and draw tangents develops spatial reasoning and accuracy — skills used by architects, engineers, and surveyors every day!
Given: Line segment AB
Required: Point P on AB such that AP:PB = m:n
Step 1: Draw a ray AX making an acute angle with AB.
Step 2: Starting from A, mark (m+n) = 5 equal arcs on AX. Label them A₁, A₂, A₃, A₄, A₅.
Step 3: Join A₅ to B.
Step 4: Through A₃ (the mth point), draw a line parallel to A₅B (using the properties of parallel lines).
Step 5: Let this parallel line meet AB at point P.
Result: AP:PB = 3:2 ✓
Justification: By Basic Proportionality Theorem, P divides AB in ratio AA₃:A₃A₅ = 3:2
Case 1 — Scale factor m/n where m < n (smaller triangle):
The new triangle is smaller than the given triangle.
Case 2 — Scale factor m/n where m > n (larger triangle):
The new triangle is larger than the given triangle.
Given: △ABC. Required: △A'B'C' ~ △ABC with A'B'/AB = 2/3
Step 1: Draw a ray BX from B making an acute angle with BC.
Step 2: Mark 3 equal arcs on BX: B₁, B₂, B₃ (taking the larger denominator).
Step 3: Join B₃C.
Step 4: Through B₂ (the numerator point), draw B₂C' ∥ B₃C meeting BC at C'.
Step 5: Through C', draw C'A' ∥ CA meeting BA at A'.
Result: △A'B'C' is the required triangle with scale factor 2/3.
When the scale factor is greater than 1, the new triangle is larger. In this case, you extend BC beyond C and BA beyond A to locate C' and A' respectively.
Example: Scale factor 3/2 → Mark 3 arcs (larger number), join B₂C, draw B₃C' ∥ B₂C where C' is on BC extended.
Given: Circle with centre O. Point P on the circle.
Step 1: Join OP.
Step 2: Extend PO beyond P to get a ray.
Step 3: Draw a line perpendicular to OP at P.
Result: This perpendicular line is the tangent at P (since tangent ⊥ radius at contact point).
Step 1: Join OP.
Step 2: Find midpoint M of OP (perpendicular bisector of OP).
Step 3: Draw a circle with M as centre and MO (= MP) as radius. This circle intersects the original circle at two points — call them A and B.
Step 4: Join PA and PB.
Result: PA and PB are the two required tangents from P. PA = PB (verified!).
The circle with diameter OP passes through A and B (since ∠OAP = ∠OBP = 90° — angle in semicircle).
Since OA ⊥ PA and OB ⊥ PB, PA and PB are tangents by the converse of "tangent ⊥ radius" theorem.