# Class 9: Triangles – Exercise 9(a)

Question 1: In the Adjoining Figure, and are two points on equal sides and of such that . Prove that .

Consider and

(given)

is common

(given)

Therefore  (by S.A.S theorem)

Hence

Question 2: If is the mid point of the hypotenuse of a right triangle , prove that .

Construct ABD such that ABD = DE

Consider and

(given)

(vertically opposite angles)

(constructed)

Therefore  (by S.A.S theorem)

and

(interior angles on the same side of the transversal are supplementary)

Therefore and are right angle triangles.

Therefore

is common

Therefore

. Hence proved.

Question 3: In a quadrilateral and bisects  . Show that . What can you say about and .

Consider   and

(given)

(given)

is common

Therefore  (by S.A.S theorem)

Question 4: Prove that is isosceles if any of the following holds: (i) Altitude bisects (ii) Median is perpendicular to the base .

(i)  If

Consider  and

is common

(altitude is perpendicular to the base)

(given)

Therefore

Therefore is isosceles triangle.

(ii) If Median is perpendicular to the base

is common

(since is the median, is the midpoint of )

Therefore

Therefore is isosceles triangle.

Question 5: In the adjoining figure and . Prove that , and .

Consider  and

is given

given

(is common)

Therefore

Therefore and

Question 6: In the adjoining figure, , and . Prove that

Given and

Therefore (by S.A.S theorem)

Question 7: In a , and and are midpoints of the sides and respectively. Prove that .

Given

Consider and

Therefore

Question 8: In the adjoining figure, is a square and is an equilateral triangle. Prove that (i) (ii)

Consider and

(equilateral triangle)

(sides of a square)

Since and

Therefore

In

Therefore

Question 9: Prove that the median of an equilateral triangle are equal.

Given:

is equilateral

Therefore

To prove:

Consider and

is an equilateral triangle)

is an equilateral triangle)

(medians bisect the opposite side. Since is an equilateral triangle the sides are equal)

Therefore

Now consider and

is an equilateral triangle)

is an equilateral triangle)

Therefore

Hence

Hence we can say that

Question 10: is a line segment. and are points on opposite sides of such that each of them is equidistant from the points and . Show that the line is perpendicular bisector of .

Consider and

(given)

(bisector)

is common

Therefore

Therefore

Similarly in and

is common

Therefore

Therefore is a perpendicular bisector of

Question 11: In the adjoining figure, and , find the ratio

Since

Also since

Now

Question 12:  In the adjoining figure, and . Prove that and .

Given:

Since

Therefore

Therefore in and , we have

is common

Therefore (A.S.A. theorem)

Hence and

Question 13: In the adjoining figure, and . Prove that and hence and \$latex \$latex .

Given:

Consider and

(Proved above)

(given)

(given)

Therefore  (A.S.A theorem)

Therefore and

Question 14: In the adjoining figure, it is given that and . Prove that .

Consider

This post first appeared on Icse Mathematics « MATHEMATICS MADE EASY FOR STUDENTS, please read the originial post: here

# Share the post

Class 9: Triangles – Exercise 9(a)

×