Chapter 10: Circles

According to the question,

AB and CD are two equal chords of a circle with centre O, intersecting each other at M.

To prove:

(i) MB = MC and

(ii) AM = MD

Proof:

AB is a chord and OE ⊥ to AB from the centre O,

Since, perpendicular from the centre to a chord bisect the chord

We get,

AE = ½ AB

Similarly,

FD = ½ CD

It is given that,

AB = CD⇒ ½ AB = ½ CD

So, AE = FD … (1)

Since equal chords are equidistance from the centre,

And AB = CD

So, OE = OF

Now, as proved, in right triangles MOE and MOF,

hyp. OE = hyp. OF [Common side]

OM = OM

ΔMOE ≅ ΔMOF

ME = MF … (2)

Subtracting equations (2) from (1), we get

AE – ME = FD – MF

⇒ AM = MD [Proved part (ii)]

Again, AB = CD [Given]

And AM = MD [Proved]

AB – AM = CD – MD [Equals subtracted from equal]

Hence, MB = MC [Proved part (i)]

According to the question,

We have,

ABCD is a trapezium in which AD||BC

Non-parallel sides AB and DC of the trapezium ABCD are equal i.e.,

AB = DC.

To prove: Trapezium ABCD is cyclic.

Construction: Draw AM and DN such that they are perpendicular on BC.

Proof: In right triangles AMB and DNC,

∠AMB = ∠DNC = 90^o

AB = DC [Given]

Since perpendicular distance between two parallel lines are same,

AM = DN

ΔAMB ≅ ΔDNC [By RHS congruence rule]

∠B = ∠C [CPCT]

And ∠1 = ∠2

∠BAD = ∠1+ 90

= ∠2 + 90

= ∠CDA

Now, in quadrilateral ABCD

∠B +∠C +∠CDA+∠BAD = 360

∠B +∠B +∠CDA+∠CDA = 360

2(∠B +∠CDA) = 360

∠B +∠CDA =180

We know that,

If any pair of opposite angles of a quadrilateral is 180^o, then the quadrilateral is cyclic.

Hence, the trapezium ABCD is cyclic.

To prove: R, D, P and Q are concyclic.

Construction: Join RD, QD, PR and PQ.

RP joins the mid-point of AB, i.e., R, and the mid-point of BC, i.e., P.

Using midpoint theorem,

RP||AC

Similarly,

PQ||AB.

So, we get,

ARPQ is a parallelogram.

So, ∠RAQ = ∠RPQ [Opposite angles of a ||gm]…(1)

ABD is a right angled triangle and DR is a median,

RA = DR and ∠1 = ∠2 …(2)

Similarly ∠3 = ∠4 …(3)

Adding equations (2) and (3),

We get,

∠1+ ∠3 = ∠2 +∠4

⇒ ∠RDQ = ∠RAQ

∠RPQ [Proved above]

Since ∠D and ∠P are subtended by RQ on the same side of it, we get the points R, D, P and Q concyclic.

Hence, R, D, P and Q are concyclic.

According to the question,

ABCD is a parallelogram.

A circle through A, B is so drawn that it intersects AD at P and BC at Q.

To prove: P, Q, C and D are concyclic.

Construction: Join PQ.

Extend side AP of the cyclic quadrilateral APQB to D.

External angle, ∠1= interior opposite angle, ∠B

Since, BA||CD and BC cuts them

∠B +∠C =180^o

Since, Sum of interior angles on the same side of the transversal = 180^o

Or ∠1+∠C =180^o

So, PDCQ is cyclic quadrilateral.

Hence, the points P, Q, C and D are concyclic.

According to the question,

Triangle ABC and l is perpendicular bisector of BC.

To prove:

Angles bisector of ∠A and perpendicular bisector of BC intersect on the circumcircle of ΔABC.

Proof:

Let the angle bisector of ∠A intersect the circumcircle of ΔABC at D.

Construction: Join BP and CP.

Since, angles in the same segment are equal

We have, ∠BAP = ∠BCP

We know that,

AP is bisector of ∠A.

Then,

∠BAP = ∠BCP = ½ ∠A …(1)

Similarly,

We have,

∠PAC = ∠PBC = ½ ∠A …(2)

From equations (1) and (2),

We have

∠BCP = ∠PBC

We know that,

If the angles subtended by two Chords of a circle at the centre are equal, the chords are equal.

So,

BP = CP

Here, P is on the perpendicular bisector of BC.

Hence, the angle bisector of ∠A and the perpendicular bisector of BC intersect on the circumcircle of ΔABC.