Question

Each of two equal circles passes through the centre of the other and the two circles intersect eachother at the points A and B. If a straight line through the point A intersects the two circles at points C and D, let us prove that ABCD is an equilateral triangle.
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Answer 1

Answer:

Given :-

• Each of two equal circles passes through the centre of the other and the two circles intersect each other at the points A and B. If a straight line through the point A intersects the two circles at points C and D.

To Prove :-

• ABCD is an equilateral triangle.

Construction :-

• A, P; A, Q; B, P; B, Q; A, B and P, Q are joined.

Proof :-

⇒ AP = BP = BQ = AQ = PQ (radius of same circle )

so, ∆ APQ, ∆ BPQ are equilateral

∴ ∠ PAQ = ∠ APQ = ∠ AQP

⇒ ∠ PBQ = ∠ BPQ = ∠ PQB =60°

Because three angle are equal and their sum is 180°

Now, ∠ APB = ∠ APQ + ∠ BPQ = 60° + 60° = 120°

and ∠ AQB = ∠ AQP + ∠ PQB = 60° + 60° =120°

i.e. ∠ APB = ∠ AQB = 120°

∠ APB is the angle at the centre and ∠ADB is the angle at the peremeter

∴ ∠ ADB =1/2 ∠ APB =1/2 ×120° = 60° i.e. ∠CDB = 60°

∠ APB and ∠ACB are the angles on the same arc

∴ ∠ ACB = ∠ APB = 120°

∴ ∠BCD = 180° - ∠ ACB = 180° - 120° = 60°

Then , In ∆ BCD ,∠ CDB = ∠ BCD =60°

∴ ∠ DBC =180° - (∠ CDB +∠ BCD ) = 180° - ( 60°+ 60° ) = 60°

It is proved that ∆ ABCD is an equilateral triangle. (PROVED)

Answer 2

Answer:

Step 1

Consider the question diagram.

Join AB.

Consider centres of BC and BD to be C1 and C2 respectively.

Since both circles have same radius we can assume BC = BD. This means angle BCD = angle ADB = x.

Similarly by symmetry we can assume angle CAB = angle DAB = y,

And angle ABC = angle ABD = z.

Step 2

In triangle ABC

Two circles with centres X and Y intersect each other at A and B. The point A is joined with S, the midpoint of XY. The perpendicular to SA through A intersects the circles at P and Q. How can I prove PA= AQ?

Two circles with centres C and D meet each other at points A and B. A straight line passes through the point A and meet the both circles at points P and Q. How can I prove that angle BPC = angle BQD?

Two circles with centres C and D meet each other at points A and B. A straight line passes through the point A and meet the both circles at points P and Q. How can I prove that,(i) angle PBQ= angle CAD,(ii) angle BPC = angle BQD?

Two equal circles pass through each other’s centre. If the radius of each circle is 5 cm, what is the length of the common chord?

Two circles intersect each other such that each circle passes through the centre of the other. If the distance between their centres is 12 cm, what is the radius of each circle?

Join AB

In triangle ABC & ABD

Angle BCA=BDA {BY same chord AB on same (equal) circle }

AB=AB common

CB=CD {opp. Side of equal angles }

So traingle ABC & ABD congruent

So , angle CAB=DAB

but Angle CAB+DAB=180

2CAB=180

CAB=90 and its angle on circumferenc

So CB is diameter similarly DB is diameter

O and O’ (centers)are mid points of CB & DB respectively

Join O ,O’

Angle OBO’=60 { equilateral triangle OB=BO’=OO’(RADIUS)}

NOW Angle CBD+BCD+CDB=180

60+2BCD=180 {BCD=CDB, Proved ealier}

2BCD=180-60

BCD=120÷2=60

So BDC=BCD=CBD=60

EQUILATERAL PROVED