JEE Mathematics Geometrical Constructions Notes Set 02

Basic Constructions Used Repeatedly

1. Perpendicular bisector of a segment: draw arcs of equal radius from both ends; the line through the two intersection points is the bisector.
2. Triangle with a given base, base angle and sum of the other two sides: mark the sum along the ray of the base angle, join its end to the other end of the base, and take the perpendicular bisector of that join; its meeting point with the ray is the third vertex.
3. Triangle with a given perimeter and base angles: draw a line of length equal to the perimeter, make half the base angles at its ends, and the meeting point of the bisecting rays is the apex of a triangle whose base is found by perpendicular bisectors of the two joins.
4. Triangle with a given base, vertical angle and altitude or median: place the vertex on the circle through the base on which the base subtends the given angle, and cut that circle with the line (or circle) given by the altitude or median.

Circle on Which a Chord Subtends a Given Angle

For a base \(BC\) and vertical angle \(A\), draw the angle \(CBP=A\) below \(BC\), draw \(BE\perp BP\), and draw the perpendicular bisector of \(BC\). Their meeting point \(O\) is the centre and \(OB\) is the radius. Every point on the major arc sees \(BC\) at angle \(A\). The radius is \(R=\dfrac{BC}{2\sin A}\).

Similar Triangles and Quadrilaterals

1. To scale a triangle by \(\dfrac mn\) with \(m2. The proof uses the basic proportionality theorem: the parallel gives \(\dfrac{AB'}{AB}=\dfrac mn\), and the second parallel \(B'C'\parallel BC\) gives similarity.
3. For a quadrilateral, the same procedure is applied along a diagonal, and then parallels to the other sides give the scaled quadrilateral.
4. Lengths scale by \(k\), areas by \(k^2\).

Tangents Without the Centre

Draw a secant \(PAB\) through the external point \(P\), extend \(AP\) to \(C\) with \(PC=PA\), draw a semicircle on \(CB\) and the perpendicular to \(CB\) at \(P\) up to the semicircle at \(D\). Then \(PD^2=PA\times PB\), which equals the square of the tangent length. Arcs of radius \(PD\) about \(P\) cut the circle at the points of contact.

Equal-Area Triangle for a Quadrilateral

Divide the quadrilateral into two triangles by a diagonal, move one vertex along a line parallel to that diagonal to reduce the quadrilateral to a triangle of the same area (triangles on the same base and between the same parallels have equal areas), and repeat on the side required.