# Important Concepts in Geometry - Kamal Lohia

• Lines and Angles

A line, mathematically, is a straight line with no end points. So it has infinite length and zero thickness.

A ray, on the other hand, is a line with one end point. And it is used to form angles. Its length is also infinite.

A line segment is a part of line with both end points. Certainly its length is finite. And in common usage, we address a line segment as line. But CAT may trap you on it. So beware. :)

How do you determine whether two lines are parallel or not?

You may answer: if they do not intersect.

Or someone may also say that if distance between them is constant.

Let’s frame it in a beautiful data sufficiency question.

Are the lines P and Q are parallel?

I. P and Q do not intersect.

II. Perpendicular distance of any point on P from Q is constant.

If you are thinking that question can be answered by using either statement alone, them I am sorry but you are wrong.  Two lines in different parallel planes will certainly never intersect. But it is not compulsory that distance between them is constant.  Now important point is that, if two lines are parallel then there is certainly a plane which passes through both of them.

Another type of problems associated with lines

How many regions/parts are created when a line is drawn in a plane?

Certainly a line will divide the plane in two parts.

Next…In how many regions/parts a plane is divided by two lines?

Now it depends on whether the two lines intersect or not. If they don’t intersect, number of parts created is 3 and if the lines intersect then number of parts is 4.

But what if question asks to find the number of parts created by n lines in a plane? :)

Clearly, as we saw earlier, number of parts (P) created depends on number of lines (L) and number of intersection (I) of two lines.  Now the relation between them (without proof) is:

P = L + I + 1

Important point to keep in mind is that I is the intersection of two lines at a point. If more than two lines pass through same point then we need to increase I by number of extra lines passing through same point. 10 chords are drawn in a circle. Find the maximum number of parts in which circle is divided.

For maximum parts, every two chords must intersect and intersect at different point.

So answer becomes, P = L + I + 1 = 10 + C (10, 2) + 1 = 56

According to measurement, angles can be classified as: There are two more classification –

Supplementary Angles – two angles whose sum is 1800 are said to be supplement of each other.

Complementary Angles – two angles whose sum is 900 are said to be complement of each other.

Now come angles formed by intersection of lines and parallels intersected by transversals. In the above figure, l and m are two parallel lines intersected by a transversal PS. The following properties of the angles can be observed:

< 1 = < 7 and < 2 = < 8 [AlternateExterior Angles]

< 3 = < 5 and < 4 = < 6 [Alternate Interior angles]

< 1 = < 5, < 2 = < 6, < 4 = < 8, < 3 = < 7 [Corresponding angles]

< 4+ < 5= < 3+ < 6=180o [Co-interior angles are interior angles on same side of the transversal that adds up to 180o]

< 1+ < 8= < 2+ < 7=180o

< 1+ < 8= < 2+ < 7=180o [Co-exterior angles are exterior angles on same side of the transversal that adds up to 180o]

Besides these, we discriminate triangles according to their angles.

Acute angled Triangle – have all three angles acute.

Right angled Triangle – have one right angle and other two acute.

Obtuse angled Triangle – have one obtuse angle and other two acute.

Also, if in a triangle two angles are equal, then sides opposite to those equal angles are also equal in length. And, if in a triangle two angles are unequal, then side opposite to smaller angle is smaller.

We can also find a direct relationship between side lengths and angles of a triangle – given by

Sine rule as below: Another important one is Cosine rule: Where a, b, c are the side lengths opposite to vertices A, B and C respectively as shown below. And R is the radius of circumscribe circle of the triangle ABC. Two basic properties related to angles in triangle:

Angle Sum Property – Sum of all the three angles of a triangle is 180o.

Exterior Angle Property – Exterior angle at a vertex of a triangle is equal to sum of two opposite angles.

Triangles

According to lengths of the side, we divide triangles in three categories.

Equilateral – all sides are equal.

Isosceles – at least two sides are equal.

Scalene – all three sides are different.

But can we form a triangle with any three sides?

Let’s talk with an example:

What is the area of the triangle whose three side lengths are 1, 2 and 3 units respectively?

Triangle is not possible. Or, mathematically, we should say area of triangle is zero.

So question arises, what is the condition on lengths of sides of a triangle to confirm that a triangle is possible.

Just lay down the longest side as base of the triangle. Now remaining two sides should not overlap on base. So, for this reason, sum of those other two side lengths of the triangle should be greater than longest side of the triangle.

Triangle Inequality: If a, b, c are the side lengths of a triangle then sum of any two sides of a triangle is greater than third side i.e. a + b > c, b + c > a and c + a > b

Also difference of two sides is always less than third side of triangle
i.e. a – b < c, b – c < a and c – a < b I know that you know almost all of these formulae already. But target here is to analyze them more deeply. If I ask you under what conditions the area of the triangle becomes maximum, can you think a way out?

Some special lines and points in triangles

MEDIAN & CENTROID

A line segment joining mid-point of a side of a triangle with opposite vertex is known as MEDIAN.

There are three medians possible in every triangle.

Median divides a triangle in two equal area triangles.

All the three medians intersect at a point, known as CENTROID.

CENTROID divides every median in the ratio 2 : 1

All the three medians divided the triangle in six equal area triangles.

CENTROID always lies inside a triangle.

ANGLE BISECTORS & INCENTER Each angle bisector of the triangle divides the opposite side in the ratio of the sides containing the angle. (Angle Bisector Theorem)

All three angle bisectors meet at a point which is known as INCENTER i.e. center of inscribed circle.

INCENTER is a point inside the triangle which is equidistant from three sides.

INCENTER is center of inscribed circle of the triangle which touches all the three sides of the triangle.

INCENTER always lies inside a triangle.

PERPENDICULAR BISECTOR & CIRCUMCENTER

Perpendicular bisector is a line which divides a side of the triangle in two equal halves perpendicularly.

All the three perpendicular bisectors of a triangle intersect at a common point which is known as CIRCUMCENTER.

CIRCUMCENTER is a point which is equidistant from three vertices of the triangle.

CIRCUMCENTER is center of circumscribed circle of the triangle which passes through all the three vertices.

For acute triangle, circumcenter lies inside the triangle.

For right triangle, circumcenter lies at mid-point of hypotenuse.

For obtuse triangle, circumcenter lies outside the triangle.

ALTITUDES & ORTHOCENTER

Perpendicular dropped from a vertex to opposite side is known as an ALTITUDE.

All the three altitudes meet at a common point known as ORTHOCENTER.

Some special triangles

EQUILATERAL TRIANGLE

All three sides are same in length (say a).

Each angle is equal and equal to 60o.

All the four lines and points discussed above are same points in an equilateral triangle.

Height of an equilateral triangle is √3 a/2

Area of an equilateral triangle is √3 a2/4

RIGHT TRIANGLE

Pythagoras Theorem states a2 + b2 = c2 where c is the hypotenuse and a, b are legs.

There are two special cases of right triangles: 30-60-90 and 45-45-90 as shown. Note the ratio of their sides. Congruent Triangles

The two triangles, whose any three corresponding parameters (out of three sides and three angles) are identical are same to be congruent.

There are three exceptions to this rule and they are AAA, SSA and ASS.

Two congruent triangles are identical in all respects.

Similar Triangles

The two triangles are said to be similar

(I) if two of the angles of triangles are same – AA, or
(II) if two sides of a triangle are proportional to two sides of another triangle and angle included between the two sides in both the triangles is same – SAS.

Sides opposite to equal sides are said to be proportional sides and ratio of proportional sides is same. Area of two similar triangles is in the ratio of square of ratio of corresponding sides.

Inradius (r) is radius of the incircle inscribed in a triangle which touches all the sides of the triangle.

Circumradius (R) is radius of circumcircle passing through all the vertices of triangle.

We have already mentioned the formulae to find r and R in terms of area of triangle.

If b, c are two sides of triangle and h is the altitude drawn from vertex common to b and c, then R = bc/2h.

For a right triangle with legs a, b and hypotenuse c, inradius r = (a + b – c)/2

If side lengths of a quadrilateral are a, b, c and d then its area is not constant. By changing internal angles, we can change area of the quadrilateral. For example, take the case of a square with side length 5 whose area is 25 square units. Now keeping the side lengths constant, if we tilt it to turn it into a rhombus, area goes o decreasing which can ultimately minimized to zero when the sides overlap with each other. So, for a given set of four side lengths, Maximum Area is obtained for a cyclic quadrilateral which is given by √ [(s - a)(s - b)(s - c)(s - d)] (BRAHMGUPTA’s formula) , where s is semiperimeter i.e. (a + b + c + d)/2.

Remember that for every four side lengths which form a quadrilateral, a cyclic quadrilateral can be formed.

If D1 and D2 are the two diagonals of a quadrilateral which intersects at angle P, then it’s area is given by ½D1D2sinP. And this area will be largest when diagonals intersect at right angle.

Ptolemy’s inequality For a quadrilateral with side lengths a, b, c, d and diagonals of length D1 and D2,
we have D1D2 < = ac + bd

Equality arises in case of a cyclic quadrilateral.

If two diagonals divide a quadrilateral in four triangles, as shown,

Whose areas are A, B, C and D in order, then

A × C = B × D

1

1

6

1

1

12

1

2