Geometry Tutorial Part - 3 - Campus Placements

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Geometry Tutorial Part - 3

Author: Sureshbala

GEOMETRIC CENTRES OF A TRIANGLE

CIRCUMCENTRE

The three perpendicular bisectors of a triangle meet at a point called circumcentre of the triangle and it is represented by S. The circumcentre of a triangle is equidistant from its vertices is called circumradius (represented by R) of the triangle. The circle drawn with the circumcentre as centre and circumradius as radius is called the circumcircle of the triangle and passes through all there vertices of the triangle.(refer to Fig. 1.13).

INCENTRE AND EXCENTRES

The internal bisectors of the three angles of a triangle meet at a point called incentre of the triangle and it is represented by I. incentre is equidistant from the three sides of the triangle i.e., the perpendicular's drawn from the in centre to the three sides are equal in length and this length is called the inradius (represented by r) of the triangle. The circle drawn with incentre as centre and inradius as radius is called the incircle of the triangle and it touches all three sides on the inside. In Fig.1.14 $\angle BIC=80^{\circ}+\cfrac {1}{2} A$ Where Iis the incentre $\angle CIA =90^{\circ}+\cfrac {1}{2} B$ and $\angle AIB=90^{\circ}+\cfrac {1}{2}C$

If the internal bisector of one angle and the external bisectors of the other two angles are drawn, they meet at a point called excentre. Three will be totally three excentres for the triangle-one corresponding to the internal bisector of each angle.

ORTHOCENTRE

The three altitudes meet at a point called Orthocenter and it is represented by O (refer to Fig. 1.15) $\angle =180^{\circ}-A, \angle COA=180^{\circ}-B, \angle AOB=180^{\circ}-C$

CENTROID

The three medians of a triangle meet at a point called the centroid and it is represented by G (Refer to Fig. 1.16).

Important points about geometric canters of a triangle

Please note the following important points pertaining to the geometric centers of a triangle ABC. In an acute angled triangle, the circumcenter lies inside the triangle. In a right-angled triangle, the circumcenter lies on the hypotenuse of the triangle (it is the midpoint of the hypotenuse). In an obtuse angled triangle, the circumcenter lies outside the triangle.

In an acute angled triangle, the orthocenter lies inside the triangle. In a right-angled triangle, the vertex where the right angle is formed (i.e., the vertex opposite the hypotenuse) is the orthocenter. In an obtuse angled triangle, the orthocenter lies outside the triangle.

In a right-angled triangle the length of the median drawn to the hypotenuse is equal to half the hypotenuse. This median is also the circumradius of the right-angled triangle.

Centroid divides each of the medians in the ratio 2:1, the part of the median towards the vertex being twice in length to the part towards the side.

Inradius is less than half of any of the three altitudes of the triangle.

In an isosceles triangle, the centroid, the orthocenter, the circumcentre and incentre, all lie on the median to the base.

In an equilateral triangle, the centroid, the orthocenter, the circumcentre and the incentre, all coincide.

Hence, in equilateral triangle ABC shown in Fig.1.17 AD is the median, altitude, angular bisector and perpendicular bisector. G is the centroid which divides the median in the ratio of 2:1. Hence, AG =2/3 and GD = 1/3 AD.

But since AD is also the perpendicular bisector and angular bisector and since G is the circumcentre as well as the incentre, AG will b the circumradius and GD will be the inradius of the equilateral triangle ABC Since AD is also the side of the equilateral triangle ABC. Since AD is also the altitude, its length is equal to $\sqrt {3}a/2$ where a is the side of the equilateral triangle. Hence, the circum radius of the equilateral $\cfrac {2}{3}*\cfrac {\sqrt{3}}{2}.a=a/ \sqrt{3}$ and the inradious $\cfrac {1}{3}*\cfrac {\sqrt{3}}{2}.a=a/ 2 \sqrt{3}$

Since the radii of the circumcircle and the incircle of an equilateral triangle are in the ratio 2 : 1, the areas of the circumcircle and the incircle of an equilateral triangle will be in the ratio of 4:1.

When the three medians of a triangle (i.e., the medians to the three sides of a triangle from the corresponding opposite vertices) are drawn, the resulting six triangles are equal in area and the area of each of these triangles in turn is equal to one-sixth of the area of the original triangle.

In Fig 1.18, AD, BE and CF are the medians drawn to the three sides. The three medians meet at the centroid G. The six resulting triangles AGF, BGF, BGD, CGD, CGE and AGE are equal in area and each of them is equal to $\cfrac {1}{6}th$ of the area of triangle ABC.

SIMILARITY OF TRIANGLES

Two triangles are said to be similar if the three angles of one triangle are equal to the three angles of the second triangle. Similar triangles are alike in shape only. The corresponding angles of two similar triangles are equal but the corresponding sides are only proportional and not equal.

For example, in Fig 1.19, if $\triangle ABC$ is similar to $\triangle DEF$ where $\angle A= \angle D, \angle B=\angle E \,and \, \angle C=\angle F$ then we have ratios o the corresponding sides equal, as given below.

$\cfrac {AB}{DE}$ $=\cfrac {BC}{EE}$ $=\cfrac {CA}{FD}$

By "corresponding sides", we mean that if we take a side opposite to a particular angle in one triangle, we should consider the side opposite to the equal angle in the second triangle. In this case, since AB is the side opposite to $\angle C \, in \, \angle ABC$ , and since $\angle C=\angle F$ , we have DE which is the side opposide to $\angle F \, in \, \triangle DEF$ .

Two triangles are similar if,

the three angles of one triangle are respectively equal to the three angles of the second triangle, or

two sides of one triangle are proportional to two sides of the other and the included angles are equal, or

if the three sides are of one are proportional to the three sides of another triangle.

In two similar triangles,

(a) Ratio of corresponding sides = Ratio of heights (altitudes) = Ratio of the lengths of the medians = Ratio of the lengths of the angular bisectors = Ratio of inradii = Ratio of circumradii = Ratio of perimeters.

(b) Ratio of areas = Ratio of squares of corresponding sides.

In a right-angled triangle, the altitude drawn to the hypotenuse divides the given triangle into two similar triangles, each of which is in turn similar to the original triangle. In triangle ABC in Fig. 1.20, ABC is a right- angled triangle where $\angle$ is a right angle. AD is triangle the perpendicular drawn to the hypotenuse BC. The triangles ABD, CAD and CBA are similar because of the equal angles given below.

In triangle ABC,

$\angle A=90^{\circ}$ if $\angle B=\theta$ , then $\angle C=90^{\circ}-\theta$ in triangle ABD , $\angle ADB=90^{\circ}$ we already know that $\angle B=\theta$ , hence $\angle BAD =90^{\circ}-\theta$

Triangle ABD, $\angle ABD=90^{\circ}$ we already know that $\angle C=90^{\circ}-\theta ,$ Hence $\angle CAD=\theta$