In GMAT, GRE, SAT, CAT and other similar Entrance Exams, the problems relating to Geometry cover mostly triangles, quadrilaterals and circles. Even though polygons with more than four sides are also covered, the emphasis on such polygons is not as much as it is on triangles and circles. In this chapter, we will look at some properties as well as theorems and riders on parallel lines, angles, triangles (including congruency and similarity of triangles), quadrilaterals, circles and polygons.
ANGELS AND LINES
An angle of is a
right angle; an angle less than
is
acute angle; an angle between
and
is an obtuse angle;
and
angle
between and is a reflex angle.
The sum of all angles made on one side of a straight line AB at a
point O by any number of lines joining the line AB at O is
. In Fig. 1.01 below, the sum of the angles
u, v, x, y and z is equal to
.
When any number of straight lines join at a point, the sum of all
the angles around that point is . In
Fig 1.02 below, the sum of the angles u, v, w, x, y, and z is
equal
.
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Two angles whose sum is are
said to be complementary angles and two angles whose sum
is are said to be supplementary angles.
When two straight lines intersect, vertically opposite angles are
equal. In Fig.1.03 given below, and
are vertically opposite angles and
and
are
vertically opposite angles. So, we have
and
Two lines which make an angle of with
each other are said to be PERPENDICULAR
to each other.
If a line passes through the mid-point of another line
then the line
is
said to be the BISECTOR of the line
i.e.,
the line
is divided into two equal parts.
If a line is drawn at the vertex of an angle dividing the angle two
equal parts, then the line
is said to be the
ANGULAR BISECTOR of the angle. Any
point on the angular bisector of an angle is EQUIDISTANT from the two arms of the angle.
If a line is perpendicular to line
as well as passes through
the mid-point of line
, then the line
is said to be
the PERPENDICULAR BISECTOR of the line
.
Any point on the perpendicular bisector of a line is EQUIDISTANT from both ends of the line.
In Fig 1.04, line PQ is the perpendicular bisector of line AB. A point P on the perpendicular bisector of AB will be equidistant from A and B, i.e., PA = PB Similarity. For any point R on the perpendicular bisector PQ, RA = RB.
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