Circle
Exercise 10.6
Part 2
Question 7: AC and BD are chords of a circle which bisect each other. Prove that (i) AC and BD are diameters, (ii) ABCD is a rectangle.
Answer: This figure shows a circle in which two chords AC and BD bisect each other and point of bisection is O
To Prove: (i) AC and BD are diameters and(ii) ABCD is a rectangle
Construction: Join AB, BC, CD and DA
Proof: (i) In ΔAOB and ΔCOD
AO = CO (O is the midpoint of AC)
BO = DO (O is the midpoint of BD)
∠AOB = ∠COD (vertically opposite angles)
So, ΔAOB ≅ΔCOD (SAS theorem)
So, AB = CD
Or, arc AB = arc D ………………(1)
Similarly, arc AD = arc BC ………(2)
Adding these equations, we get
Arc AB + arc AD = arc CD + arc BC
So, BD divides the circle into two equal parts
Hence, BD is a diameter
Similarly, AC is a diameter
Now, let us solve question (ii).
We have proved ΔAOB ≅ δCOD
Or, ∠OAB = ∠OCD
Or, ∠CAB = ∠ACD
Or, AB||DC Similarly, AD||BC can be proved
This proves that ABCD is a parallelogram
∠DAB = ∠DCB (Opposite angles of parallelogram are equal)
But ∠DAB + ∠DCB = 180° (Opposite angles of cyclic quadrilateral are complementary)
So, ∠DAB = 90° = ∠DCB
So, it is proved that ABCD is a rectangle
Question 8: Bisectors of angles A, B and C of a triangle ABC intersect its circumcircle at D, E and F respectively. Prove that the angles of the triangle DEF are `90°-1/2A`, `90°-1/2B` and `90°-1/2C`
Answer: This figure shows a triangle ABC inscribed in a circle and bisectors of ∠A, ∠B and ∠C intersect the circumcircle at D, E and F respectively.
Construction: Join DE, EF and FD
Proof: Since angles in the same segment are equal
So, ∠FDA = ∠FCA ………………..(1)
∠EDA = ∠EBA ……………………..(2)
Adding these equations, we get
∠FDA + ∠EDA = ∠FCA + ∠EBA
`=1/2∠C+1/2∠B`
`=1/2(∠C+∠B)=1/2(180°-∠A)`
`=90°-(∠A)/2`
Similarly, ∠FED = `(90°-(∠B)/2)`
And, ∠EFD = `(90°-(∠C)/2)`
So, angles of ΔDEF are as follows:
`(90°-(∠A)/2)`, `(90°-(∠B)/2)`, and `(90°-(∠C)/2)`
Question 9: Two congruent circles intersect each other at points A and B. Through A any line segment PAQ is drawn so that P, Q lie on the two circles. Prove that BP = BQ.
Answer: This figure shows two congruent circles intersecting each other at points A and B. A line segment passing through A, meets the circles at points P and Q. Let us draw the common chord AB
∠APB = ∠AQB (angles subtended by equal chords in congruent circles)
In ΔPBQ
∠AQB = ∠APB
So, BP = BQ proved
Question 10: In any triangle ABC, if the angle bisector of ∠A and perpendicular bisector of BC intersect, prove that they intersect on the circumcircle of the triangle ABC.
Answer: This figure shows ΔABC and O is the centre of its circumcircle. Perpendicular bisector of BC passes through O and intersects the circle at P. Now, join OB and OC
We need to prove that AP is the bisector of ∠BAC
Let us assume that arc BC makes angle θ at circumference
Then ∠BOC = 2θ
(Because angle at centre is double the angle made by an arc at circumference)
Also, in ΔBOC
OB = OC (Radii)
OP is perpendicular bisector of BC
So, ∠BOP = ∠COP = θ
Arc CP makes angle θ at O
So, ∠PAC = `1/2` θ
(Because angle at centre is double the angle made by an arc at circumference)
This means that AP is the bisector of ∠BAC