Question 1: If two isosceles triangles have a common base, prove that the line joining their vertices bisects them at right angles.

Answer:

Given: and

To prove: bisects at right angles

Proof: In and we have

(Given)

(Given)

and is common

Therefore (By SSS theorem))

Thus in and , we have

(Given) and (By SAS theorem)

and

But we know,

Hence, bisects at right angles.

Question 2: and are two isosceles triangles on the same base and vertices and are on the same side of . If is extended to intersect at , show that (i) (ii) (iii) bisect as well as

Answer:

(i) In and , we have

(Given)

(Given)

and is common

Therefore (By SSS theorem)

(ii) In and

(Given)

(corresponding angles of congruent triangles are equal)

and is common.

Therefore, (by SAS theorem)

(iii) In (i) we proved

is the bisector of

In and we have

(Given)

(corresponding sides of congruent triangles)

and is common

Therefore (By SSS theorem)

is the bisector of

Hence is the bisector of as well as

(iv) In (iii) we proved that,

and

and

is the perpendicular bisector of

Question 3: A point is taken inside an equilateral four sides figure such that its distances from the angular points and are equal. Show that and are in one and in the same straight line.

Answer:

Given:

To Prove: and are in one and the same straight line

Proof: In and , we have

(given)

is common

(given)

(by SSS theorem)

Similarly,

But right angles

and form a linear pair

and are in same straight line

is a straight line

Question 4: In the adjoining figure, it is given that and . Prove that .

Answer:

Consider and

(given)

(given)

is common

(By SSS theorem)

Question 5: In , if and and are the midpoints of the sides and respectively. Prove that

Answer:

In

(given)

and (given)

To prove:

Using the Mid Point theorem

and

Similarly

In and

is common

(by SSS theorem)

Question 6: In the adjoining figure, it is given that and and . Prove that . Hence prove that the triangles is isosceles.

Answer:

Consider and

(given)

(given)

and

Question 7: is a square, and are points on the side and respectively such that . Prove that and

Answer:

Consider and

is common

(given)

and

Question 8: In the adjoining figure, . and are the bisectors of and respectively. Prove that latex > DC $.

Answer:

In

(given)

(angle opposite to the larger side of a triangle is greater)

and

Question 9: In the adjoining figure, side and of are extended to and respectively. If , show that .

Answer:

We have

(angles of a linear pair)

(angles of a linear pair)

Since

Question 10: If is any point on the base produced, of an isosceles triangle , prove that .

Answer:

In

(given)

(angles opposite equal sides of a triangle are equal)

In , we have

Exterior … … … … … i)

(since exterior angle of a triangle is greater than each of interior opposite angles)

… … … … …. ii)

From i) and ii), we get

Question 11: In the adjoining figure, if is the bisector of , show that (i) (ii)

Answer:

In ,

is the bisector of

(since exterior angle of a triangle is greater than each of interior opposite angles)

Therefore in

Thus in we have

(sides opposite to the greater angle is larger)

Hence .

Similarly, we can prove .

Question 12: In the adjoining figure, and is the bisector of . Show that .

Answer:

In , we have

Since

Now in and we have

and

or

Question 13: Show that the sum of three altitudes of a triangle is less than the sum of the three sides of the triangle.

Answer:

Given (altitudes)

and

… … … … … i)

Similarly,

and

… … … … … ii)

and

and

… … … … … iii)

Adding i) , ii) and iii) we get

Question 14: Prove that the perimeter of a triangle is greater than the sum of its three medians.

Answer:

In and are medians.

To prove:

Proof: We know that the sum of any two sides of a triangle is greater then the twice of the median bisecting the third side.

is the median bisecting

Similarly,

and

Adding i), ii) and iii) we get

Question 15: In the adjoining figure, is a triangle and is any point in its interior, show that

Answer:

Given is any point inside

To Prove:

Proof: In we have

(Since the sum of two sides of a triangle is greater than the third side)

… … … … … i)

In , we have

… … … … … ii)

Adding i) and ii) we get,

Question 16: In the adjoining figure, and . Show that .

Answer:

Given

In and , we have

is common

and (by construction)

(by SAS theorem)

(since corresponding part of congruent triangles)

Thus in , we have

… … … … … i) (angles opposite equal sides are equal)

In , we have

… … … … … ii)

from i) and ii) we get

(since side opposite to greater angle is larger)

Hence

Question 17: In the adjoining figure, is a quadrilateral. is the longest side and is the shortest side. Prove that and

Answer:

To prove: i) ii)

Construction: Join and

Since is the longest side of Quadrilateral , therefore in , we have

… … … … … i)

Since is the shortest side of the quadrilateral , therefore we have in

… … … … … ii)

Adding i) and ii) we get

ii) In , we have

… … … … … iii)

In , we have

… … … … … iv)

Adding iii) and iv) we get

Hence Proved.

Question 18: In the adjoining figure, is a quadrilateral in which diagonals and intersect at . Show that (i) (ii)

Answer:

i) Since the sum of any two sides of a triangle is greater than the third side, therefore

In we have

… … … … … i)

In we have

… … … … … ii)

In , we have

… … … … … iii)

In , we have

… … … … … iv)

Adding i), ii), iii) and iv) we get

In , we have

… … … … … v)

In , we have

… … … … … vi)

In , we have

… … … … … vii)

In , we have

… … … … … viii)

Adding v), vi), vii) and viii) we get

Since and

Question 19: is any point in the interior of . Prove that (i) (ii) (iii)

Answer:

In we have

… … … … … i)

In , we have

… … … … … ii)

Adding i) and ii), we get

… … … … … iii)

ii) Similarly,

… … … … … iv)

and … … … … … v)

Adding iii) , iv) and v) we get

In and , we have

Question 20: In the adjoining figure, prove that: (i) (ii)

Answer:

i) In we have

… … … … … i)

In we have

… … … … … ii)

Adding i) and ii) we get

ii) In we have

(since )