Geometric construction is the method of accurately drawing geometric shapes using basic tools like a compass and a ruler, based on mathematical rules and properties of lines, angles, and circles.
Fundamental Geometric Constructions
The basic constructions are:
Angle Bisector: A line that cuts an angle in half exactly, always going through the corner point (vertex). These cuts are important for solving geometry problems.
Tangent to a circle: A line that touches a circle at exactly one point (the point of contact) and is perpendicular to the radius at that point.
Parallel Lines: Two lines on the same flat surface (plane) that will always stay the same distance apart and never cross.
Copying an Angle: Making a new angle that has the same size (measure) as another angle, but in a different location.
Construction of a Perpendicular Bisector
To carry out this construction, we use the fact that any point on the perpendicular bisector of a line segment is equidistant from its two endpoints. Suppose we have a line segment AB.
Steps of Construction
Step 1: Taking A and B as centres and a radius greater than half of AB, draw arcs on both sides of the line segment so that they intersect each other. Step 2: Let the points of intersection be P and Q. Join P and Q to form a line. This line is the perpendicular bisector of AB.
To construct an angle bisector, we use the fact that it divides an angle into two equal parts. Suppose we have an angle ∠PQR and we want to bisect it.
Steps of Construction
Step 1: With Q as the centre and any convenient radius, draw an arc that intersects the rays QP and QR at points E and D, respectively. Step 2: With E and D as centres and using the same radius, draw arcs that intersect each other at point F. Step 3: Join Q to F and draw the ray QF.
Ray QF is the angle bisector of ∠PQR, dividing it into two equal angles.
Basic Proportionality Theorem (BPT)
The Basic Proportionality Theorem states that if a line is drawn parallel to one side of a triangle, it divides the other two sides in the same ratio. In triangle ABC, if a line DE is drawn parallel to BC and intersects AB at D and AC at E, then:
AD/DB = AE/EC
An angle in a Semicircle is 90°
The angle subtended by a diameter of a circle at any point on the circle is always a right angle. In a circle with centre O, if AB is the diameter and C is any point on the circle, then ∠ACB = 90°.
Scale Factor
The scale factor is the ratio of the corresponding sides of the new triangle to the given triangle. It shows how much the new triangle is enlarged or reduced compared to the original triangle.
Applications of Geometric Constructions
This section includes three main geometric constructions:
Constructing tangents to a circle from an external point.
Dividing a Line in a Given Ratio
A line segment can be divided into a given ratio using geometric construction without measuring its length directly.
Example: Divide a line segment AB in the ratio 4:5.
Steps of Construction
Step 1: Draw a line segment AB. From point A, draw a ray AX making an acute angle with AB. Step 2: Mark 5 equal points (A₁, A₂, A₃, A₄, A₅) on ray AX using a compass. Step 3: Join A₅ to B. Through point A₄, draw a line parallel to A₅ B. Let this line intersect AB at C.
Point C divides AB in the ratio 4a:5, i.e., AC / CB = 4 / 5.
In triangle ABA₅, by the Basic Proportionality Theorem,
AC / CB = AA₄ / A₄A₅ = 4 / 5.
Construct a Similar Triangle with the Given Scale Factor
A triangle similar to a given triangle can be constructed using a specified scale factor.
Example: Construct a triangle similar to △ABC with a scale factor of 4:5.
Steps of Construction:
Step 1: Draw a triangle ABC. From point B, draw a ray BX making an acute angle with BC. Step 2: Mark 5 equal points (B₁, B₂, B₃, B₄, B₅) on ray BX using a compass. Step 3: Join B₅ to C. Through point B₄, draw a line parallel to B₅C. Let it intersect BC at C′. Step 4: Through point C′, draw a line parallel to AC. Let it intersect AB at B′. Triangle AB′C′ is the required similar triangle.
Point B′C′ forms a triangle similar to ABC in the ratio 4: 5.
In triangle BB₅C, by the Basic Proportionality Theorem,
BC′ / C′C = BB₄ / BB₅ = 4 / 5
In triangle ABC, again by BPT,
BB′ / B′A = BC′ / C′C = 4 / 5
Construct Tangents of a Circle from an External Point
From a point outside a circle, exactly two tangents can be drawn to the circle.
Example: Construct tangents from a point P to a circle of radius 4 cm, where the distance of P from the centre O is 10 cm.
Steps of Construction
Step 1: Draw a circle with centre O and radius 4 cm. Mark an external point P such that OP = 10 cm. Join OP, intersecting the circle at Q. Step 2: Construct the perpendicular bisector of OP. Let it intersect OP at M. Step 3: With M as the centre and radius MO, draw a circle. It intersects the given circle at points C and D. Join PC and PD.
PC and PD are the required tangents to the circle.
Explanation:
Join OC. Since C lies on the circle with diameter OP, ∠PCO = 90° (angle in a semicircle).
A tangent is perpendicular to the radius at the point of contact, hence PC and PD are tangents.
