Question 1: In an equilateral triangle, prove that the centroid and the center of the circum-circle (circum-center) coincide.

Answer:

Refer to the adjoining diagram.

Given: is an equilateral triangle. and are mid points of and respectively.

To Prove: Centroid and Circum-center are coincident.

Construction: Draw medians and

Proof: Consider and

(given)

BC is common

Since and are mid points of and respectively, and because is equilateral we have

Therefore (By S.A.S criterion)

… … … … … i)

Similarly,

… … … … … ii)

From i) and ii) we get

is equidistant from the vertices

is the circum-center of

Question 2: Two circles whose centers are and intersect at . Through , a line intersecting the circles at and is drawn. Prove that .

Answer:

Refer to the adjoining diagram.

Construction: Draw and

Proof:

… … … … … i)

(perpendicular drawn from the center of the circle bisect the chord)

Similarly,

… … … … … ii)

Therefore

Question 3: Prove that the line joining the mid points of two parallel chords of a circle passes through the center.

Answer:

Refer to the adjoining diagram.

Construction: Join and . Draw and

Proof: Since is the mid point of

(Theorem 4)

Since

Therefore

Similarly,

(Theorem 4)

Since

Therefore

Hence

Hence is a straight line.

Question 4: In the adjoining figure , Prove that

Answer:

Refer to the adjoining diagram.

Since congruent arcs of a circle subtend equal angles at the center

Therefore

Consider and

(radius)

(radius)

Therefore (By S.A.S criterion)

Hence

Question 5: If two chords of a circle are equally inclined to the diameter through their point of intersection, prove that the chords are equal.

Answer:

Refer to the adjoining diagram.

Given:

Construction: Draw and

To Prove:

Proof: Consider and

is common

(given)

Therefore (By A.A.S criterion)

Hence

Since equidistant chords are equal,

Question 6: In adjoining figure, is the center of a circle and bisects . Prove that .

Answer:

Refer to the adjoining diagram.

To Prove:

Construction: Draw and

Proof: Consider and

is common

( bisects – given)

Therefore

and are equidistant

Hence (equidistant chords in a circle are equal)

Question 7: Two equal chords and of a circle with center , when produced meet at a point as shown in the adjoining diagram. Prove that and .

Answer:

Refer to the adjoining diagram.

Given:

Construction: Draw and

To Prove: and

Proof: Consider and

is common

Since

they are equidistant

Therefore

Hence (By S.A.S criterion)

Therefore … … … … … i)

We know … … … … … ii)

i) – ii) we get

.

Since and ,

. Hence proved

Question 8: Prove that the line joining the mid points of two equal chords of a circle subtend equal angles with the chords.

Answer:

Refer to the adjoining diagram.

Given: and are mid points of and respectively.

To Prove: and

Construction: Draw and

Proof: (equal chords in a circle are equidistant)

Therefore

Similarly,

Question 9: In adjoining figure, and are mid points of two equal chords and of a circle with center . Prove that i) ii)

Answer:

Refer to the adjoining diagram.

Given: and

Proof: Since equal chords are equidistant from the center,

In

(angle opposite equal sides of a triangle are equal)

Question 10: and are chords of a circle equidistant from the center. Prove that the diameter passing through bisects and .

Answer:

Refer to the adjoining diagram.

Given: and are equidistant from

To Prove:

( bisects )

( bisects )

Construction: Join and

Proof: Equidistant chords are equal

Consider and

is common

(angles subtended by diameter)

(By. S.A.S criterion)

and . Hence proved.

Question 11: If two chords of a circle bisect each other, show that they must be diameters.

Answer:

Refer to the adjoining diagram.

Construction: Join and .

To Prove: and are diameters

Proof: Consider and

(Mid point of )

(Mid point of )

(vertically opposite angles)

(By S.A.S criterion)

… … … … … i)

Now consider and

(Mid point of )

(Mid point of )

(vertically opposite angles)

… … … … … ii)

Adding i) and ii)

divides the circle in 2 half’s

is a diameter

Similarly,

Therefore is diameter