Construction.

Dividing a Line Segment

Division of a Line Segment

1) Bisecting a Line Segment

Step 1: With a radius more than half the length of the line-segment, draw arcs centred at either ends of the line segment so that they intersect on either sides of the line segment.

Step 2: Join the points of intersection. The line segment is bisected by the line segment joining the points of intersection.

PQ is the perpendicular bisector of AB

2) Given a line segment AB, divide it in the ratio m:n, where both m and n are positive integers.

Suppose we want to divide AB in the ratio 3:2 (m=3, n=2)

Step 1: Draw any ray $\overrightarrow{AX}$, making an acute angle with $\overline{AB}$. 

Step 2: Locate 5 (= m + n) points $A_1,A_2,A_3,A_{4}and{A}_5$ on AX such that $A{A}_1=A_1{A}_2=A_2{A}_3=A_3{A}_4=A_4{A}_5$

Step 3: Join $B{A}_5.(A_{(m+n)}=A_5)$

Step 4: Through the point $A_3(m=3)$, draw a line parallel to $B{A}_5$ (by making an angle equal to $\angle A{A}_{5}B)$ at $A_3$ intersecting AB at the point C.

Then, AC : CB = 3 : 2.

Division of a line segment

Constructing Similar Triangles

Constructing a Similar Triangle with a scale factor

Suppose we want to construct a triangle whose sides are $\frac{3}{4}$ times the corresponding sides of a given triangle

Caption

Step 1: Draw any ray $\overrightarrow{BX}$ making an acute angle with side $\overline{BC}$  (on the side opposite to the vertex A).

Step 2: Mark 4 consecutive distances(since the denominator of the required ratio is 4) on $\overline{BX}$  as shown.

Step 3: Join $B_{4}C$ as shown in the figure.

Step 4: Draw a line through $B_3$ parallel to $B_{4}C$ to intersect BC at C'.

Step 5: Draw a line through C' parallel to AC to intersect AB at  A'. $\Delta A^{\prime }B{C}^{\prime }$ is the required triangle.

Same procedure can be followed when scale factor > 1. 

Drawing Tangents to a Circle

Tangents: Definition

A tangent to a circle is a line which touches the circle at exactly one point.

For every point on the circle, there is a unique tangent passing through it.

PQ is the tangent, touching
the circle at A

No. of Tangents to a circle from a given point

i) If the point in interior region of the circle, any line through that point will be a secant. So, no tangent can be drawn to a circle passing through a point lying inside it.

AB is a secant drawn
through the point S

ii) There is one and only one tangent to a circle passing through a point lying on the circle.

PQ is the tangent touching
the circle at A

iii) There are exactly two tangents to a circle through a point lying outside the circle.

$P{T}_1$ and (\PT_2\) are tangents
touching the circle at $T_1$ and $T_2$
 

Drawing tangents to a circle from a point outside the circle

To construct the tangents to a circle from a point outside it.

Consider a circle with center O and let P be the exterior point from which the tangents to be drawn.

Step 1: Join $\overline{PO}$ and bisect it. Let M be the midpoint of $\overline{PO}$.

Step 2: Taking M as centre and MO(or MP) as radius, draw a circle. Let it intersect the given circle at the points Q and R.

Step 3: Join PQ and PR

Step 3: $\overline{PQ}$ and $\overline{PR}$ are the required tangents to the circle.

Drawing Tangents to a circle from a point on the circle

To draw a tangent to a circle through a point on it.

Step 1: Draw the radius of the circle through the required point.
Step 2: Draw a line perpendicular to the radius through this point. This will be tangent to the circle.