Circles

Given: Two congruent circles with centres O and O′ and equal radii r. AB and CD are equal chords of the two circles respectively, i.e., AB = CD.

To Prove: ∠AOB = ∠CO′D

Proof:

In △AOB and △CO′D:

$$OA = O'C = r \quad \text{(radii of congruent circles)}$$

$$OB = O'D = r \quad \text{(radii of congruent circles)}$$

$$AB = CD \quad \text{(given)}$$

By SSS congruence criterion:
$$\triangle AOB \cong \triangle CO'D$$

Therefore, by CPCT:
$$\angle AOB = \angle CO'D$$

Hence proved. Equal chords of congruent circles subtend equal angles at their centres.

Given: Two congruent circles with centres O and O′ and equal radii r. Chords AB and CD subtend equal angles at their respective centres, i.e., ∠AOB = ∠CO′D.

To Prove: AB = CD

Proof:

In △AOB and △CO′D:

$$OA = O'C = r \quad \text{(radii of congruent circles)}$$

$$OB = O'D = r \quad \text{(radii of congruent circles)}$$

$$\angle AOB = \angle CO'D \quad \text{(given)}$$

By SAS congruence criterion:
$$\triangle AOB \cong \triangle CO'D$$

Therefore, by CPCT:
$$AB = CD$$

Hence proved. If chords of congruent circles subtend equal angles at their centres, then the chords are equal.

Given: Two circles with centres O and O′, radii 5 cm and 3 cm respectively. They intersect at points A and B. Distance OO′ = 4 cm.

To Find: Length of common chord AB.

Solution:

Let the common chord AB intersect OO′ at point M.

Let OM = x, so O′M = 4 − x.

In △OMA (right-angled at M, since the line joining centres is perpendicular to the common chord):
$$OA^2 = OM^2 + AM^2$$
$$25 = x^2 + AM^2 \quad \cdots (1)$$

In △O′MA (right-angled at M):
$$O'A^2 = O'M^2 + AM^2$$
$$9 = (4-x)^2 + AM^2 \quad \cdots (2)$$

Subtracting (2) from (1):
$$25 - 9 = x^2 - (4-x)^2$$
$$16 = x^2 - 16 + 8x - x^2$$
$$16 = 8x - 16$$
$$8x = 32$$
$$x = 4 \text{ cm}$$

Substituting in (1):
$$AM^2 = 25 - 16 = 9$$
$$AM = 3 \text{ cm}$$

Since M is the midpoint of AB (perpendicular from centre bisects the chord):
$$AB = 2 \times AM = 2 \times 3 = 6 \text{ cm}$$

Note: Since x = 4 cm = OO′, point M coincides with O′. This means O′ lies on AB, confirming AB passes through O′.

$$\boxed{AB = 6 \text{ cm}}$$

Given: A circle with centre O. Two equal chords AB and CD intersect at point E inside the circle, i.e., AB = CD.

To Prove: AE = CE and BE = DE (corresponding segments are equal).

Construction: Draw OM ⊥ AB and ON ⊥ CD.

Proof:

Since AB = CD and equal chords are equidistant from the centre:
$$OM = ON \quad \cdots (1)$$

Also, since perpendicular from centre bisects the chord:
$$MB = \frac{AB}{2} \quad \text{and} \quad ND = \frac{CD}{2}$$

Since AB = CD:
$$MB = ND \quad \cdots (2)$$

In △OME and △ONE:
$$OM = ON \quad \text{(from (1))}$$
$$\angle OME = \angle ONE = 90°$$
$$OE = OE \quad \text{(common)}$$

By RHS congruence:
$$\triangle OME \cong \triangle ONE$$

By CPCT:
$$ME = NE \quad \cdots (3)$$

Now:
$$BE = MB - ME = ND - NE = DE$$

Also:
$$AE = AB - BE = CD - DE = CE$$

Hence proved. AE = CE and BE = DE.

Given: A circle with centre O. Two equal chords AB and CD intersect at point E inside the circle. OE is joined.

To Prove: ∠OEM = ∠OEN, i.e., OE makes equal angles with the two chords (where M and N are feet of perpendiculars from O to AB and CD respectively).

Construction: Draw OM ⊥ AB and ON ⊥ CD.

Proof:

Since AB = CD (equal chords of the same circle), they are equidistant from the centre:
$$OM = ON \quad \cdots (1)$$

In △OME and △ONE:
$$OM = ON \quad \text{(from (1))}$$
$$\angle OME = \angle ONE = 90°$$
$$OE = OE \quad \text{(common)}$$

By RHS congruence:
$$\triangle OME \cong \triangle ONE$$

By CPCT:
$$\angle OEM = \angle OEN$$

Hence proved. The line joining the point of intersection to the centre makes equal angles with the chords.

Given: Two concentric circles with common centre O. A line intersects the inner circle at B and C, and the outer circle at A and D.

To Prove: AB = CD

Construction: Draw OM perpendicular to the line AD from centre O.

Proof:

For the outer circle, OM ⊥ AD:
Since perpendicular from centre bisects the chord:
$$AM = MD \quad \cdots (1)$$

For the inner circle, OM ⊥ BC:
Since perpendicular from centre bisects the chord:
$$BM = MC \quad \cdots (2)$$

Subtracting (2) from (1):
$$AM - BM = MD - MC$$
$$AB = CD$$

Hence proved.

Given: A circle of radius 5 m. Let Reshma, Salma and Mandip be at points R, S and M on the circle. RS = SM = 6 m.

To Find: RM

Solution:

Let O be the centre of the circle. Draw OA ⊥ RS and OB ⊥ SM.

Since OA ⊥ RS and perpendicular from centre bisects the chord:
$$AS = \frac{RS}{2} = 3 \text{ m}$$

In △OAS:
$$OA^2 = OS^2 - AS^2 = 25 - 9 = 16$$
$$OA = 4 \text{ m}$$

Similarly, OB = 4 m (since SM = 6 m and radius = 5 m).

Let O be the origin. Place S at the top. Since RS = SM = 6 m and both chords are equal, by symmetry, R and M are symmetric about the line OS.

Let the perpendicular from O to RS meet RS at A, and from O to SM meet SM at B.

Using coordinate geometry: Let O = (0, 0).

Since OA ⊥ RS and OA = 4 m, place A along the y-axis direction.

Let S = (0, 5) (topmost point).

For chord RS: midpoint A is at distance 4 from O along OS direction.
$$A = (0, 4)$$

R and S are symmetric about A:
$$R = (-3, 4), \quad S = (3, 4)$$

Wait — let me redo with S as a general point.

Let S = (0, 5). Chord RS has length 6, midpoint A satisfies OA ⊥ RS.

Since RS is horizontal (by symmetry, if S is at top and R, M are symmetric):
$$A = (0, 4), \quad R = (-3, 4), \quad M = (3, 4)$$

Verify: $OR = \sqrt{9+16} = 5$ ✓, $OM = \sqrt{9+16} = 5$ ✓

Check SM: $SM = \sqrt{(3-0)^2+(4-5)^2} = \sqrt{9+1} = \sqrt{10} \neq 6$

Let me use a proper coordinate approach.

Let O = (0,0). Let S = (s_x, s_y) on the circle, so $s_x^2 + s_y^2 = 25$.

Let the perpendicular from O to RS have foot A. Since OA = 4 and AS = 3:

Place the coordinate system so that S = (0, 5).

For chord RS (length 6): The perpendicular from O to RS has length $\sqrt{25-9} = 4$.

Let RS be along a line. The foot of perpendicular from O to RS is at distance 4 from O.

Let A = foot on RS. Then $\vec{OA} \perp RS$ and $|OA| = 4$, $|AS| = 3$.

So $A = S - 3\hat{u}$ where $\hat{u}$ is unit vector along RS.

Also $\vec{OA} \cdot \hat{u} = 0$, meaning $\vec{OS} \cdot \hat{u} = 3$.

Let $\hat{u} = (\cos\theta, \sin\theta)$. Then $s_x\cos\theta + s_y\sin\theta = 3$.

Similarly for chord SM: $s_x\cos\phi + s_y\sin\phi = 3$ where $\hat{v} = (\cos\phi, \sin\phi)$ is direction of SM.

This is getting complex. Let me use a direct approach.

Let O = (0,0), S = (0, 5).

For chord RS of length 6: perpendicular from O to RS = 4.
Let R = $(x_1, y_1)$ on circle, $x_1^2+y_1^2=25$, $RS=6$.

Midpoint of RS = $\left(\frac{x_1}{2}, \frac{y_1+5}{2}\right)$, and this is perpendicular to RS from O.

Distance from O to midpoint = 4:
$$\frac{x_1^2}{4} + \frac{(y_1+5)^2}{4} = 16$$
$$x_1^2 + y_1^2 + 10y_1 + 25 = 64$$
$$25 + 10y_1 + 25 = 64$$
$$10y_1 = 14 \Rightarrow y_1 = 1.4$$

Then $x_1^2 = 25 - 1.96 = 23.04$, so $x_1 = \pm 4.8$.

By symmetry, $R = (-4.8, 1.4)$ and $M = (4.8, 1.4)$.

Verify SM: $SM = \sqrt{(4.8)^2+(1.4-5)^2} = \sqrt{23.04+12.96} = \sqrt{36} = 6$ ✓

Now:
$$RM = \sqrt{(4.8-(-4.8))^2 + 0^2} = \sqrt{(9.6)^2} = 9.6 \text{ m}$$

$$\boxed{RM = 9.6 \text{ m}}$$

Given: A circular park of radius 20 m. Three boys Ankur (A), Syed (S) and David (D) sit at equal distances on the boundary, forming an equilateral triangle inscribed in the circle of radius 20 m.

To Find: Side length of the equilateral triangle (length of string).

Solution:

Since the three boys are at equal distances, they form an equilateral triangle inscribed in the circle.

For an equilateral triangle inscribed in a circle of radius R, the relationship between side $a$ and circumradius R is:
$$R = \frac{a}{\sqrt{3}}$$

Derivation using perpendicular from centre:

Let O be the centre. Draw OM ⊥ AS where M is the midpoint of AS.

Let side of equilateral triangle = $a$, so $AM = \frac{a}{2}$.

In △OAS, O is the circumcentre. The centroid divides the median in ratio 2:1.

Median of equilateral triangle $= \frac{\sqrt{3}}{2}a$

Circumradius $= \frac{2}{3} \times \frac{\sqrt{3}}{2}a = \frac{a}{{\sqrt{3}}}$

So:
$$20 = \frac{a}{\sqrt{3}}$$
$$a = 20\sqrt{3} \text{ m}$$

$$\boxed{\text{Length of each string} = 20\sqrt{3} \approx 34.64 \text{ m}}$$

Given: ∠BOC = 30°, ∠AOB = 60°. D is a point on the circle on the arc other than arc ABC.

To Find: ∠ADC

Solution:

$$\angle AOC = \angle AOB + \angle BOC = 60° + 30° = 90°$$

By the theorem — the angle subtended by an arc at the centre is double the angle subtended at any point on the remaining part of the circle:

$$\angle AOC = 2 \times \angle ADC$$

$$\angle ADC = \frac{\angle AOC}{2} = \frac{90°}{2} = 45°$$

$$\boxed{\angle ADC = 45°}$$

Given: Chord AB = radius of the circle (= r).

To Find: Angle subtended at a point on the minor arc and at a point on the major arc.

Solution:

Let O be the centre. Since OA = OB = AB = r, triangle OAB is equilateral.

$$\angle AOB = 60°$$

Angle at a point on the major arc:

Let P be a point on the major arc. By the inscribed angle theorem:
$$\angle APB = \frac{1}{2} \angle AOB = \frac{1}{2} \times 60° = 30°$$

Angle at a point on the minor arc:

Let Q be a point on the minor arc. AQBP is a cyclic quadrilateral (all on the circle). The sum of opposite angles:
$$\angle AQB + \angle APB = 180°$$
$$\angle AQB = 180° - 30° = 150°$$

$$\boxed{\text{Angle on major arc} = 30°, \quad \text{Angle on minor arc} = 150°}$$

Given: P, Q, R are on a circle with centre O. ∠PQR = 100°.

To Find: ∠OPR

Solution:

Since PQRS is not given, we use the reflex angle concept.

The angle subtended by arc PR at the centre = 2 × angle subtended at any point on the major arc.

Here ∠PQR = 100° is the angle in the major segment (Q is on the major arc).

Reflex ∠POR = 2 × ∠PQR = 2 × 100° = 200°

Therefore:
$$\angle POR = 360° - 200° = 160°$$

In △OPR:
$$OP = OR \quad \text{(radii)}$$

So △OPR is isosceles:
$$\angle OPR = \angle ORP = \frac{180° - \angle POR}{2} = \frac{180° - 160°}{2} = \frac{20°}{2} = 10°$$

$$\boxed{\angle OPR = 10°}$$

Given: ∠ABC = 69°, ∠ACB = 31°. Points A, B, C, D lie on a circle.

To Find: ∠BDC

Solution:

In △ABC:
$$\angle BAC = 180° - \angle ABC - \angle ACB = 180° - 69° - 31° = 80°$$

Angles ∠BAC and ∠BDC are angles in the same segment (both subtended by arc BC on the same side).

By the theorem — angles in the same segment are equal:
$$\angle BDC = \angle BAC = 80°$$

$$\boxed{\angle BDC = 80°}$$

Given: ∠BEC = 130°, ∠ECD = 20°. A, B, C, D are concyclic.

To Find: ∠BAC

Solution:

In △ECD (where E is intersection of diagonals AC and BD):

∠DEC = 180° − ∠BEC = 180° − 130° = 50° (linear pair)

In △ECD:
$$\angle EDC = 180° - \angle DEC - \angle ECD = 180° - 50° - 20° = 110°$$

Wait — ∠EDC = ∠BDC (same angle).

$$\angle BDC = 110°$$

Hmm, that seems large. Let me reconsider.

Actually in △EBC:
∠EBC + ∠BCE + ∠BEC = 180°

Let me use exterior angle approach.

In △BCE:
∠BEC = 130°, so ∠CBE + ∠BCE = 50°.

Using the exterior angle: In △ECD:
∠BEC is exterior angle to △ECD at E:
$$\angle BEC = \angle EDC + \angle ECD$$
$$130° = \angle EDC + 20°$$
$$\angle EDC = 110°$$

But ∠EDC = ∠BDC = 110°. This is the angle subtended by arc BC at D.

Angles ∠BAC and ∠BDC are in the same segment:
$$\angle BAC = \angle BDC$$

Wait — ∠BAC and ∠BDC subtend the same chord BC but they need to be on the same arc.

Actually, ∠BAC subtends arc BC (not containing A) and ∠BDC subtends arc BC (not containing D). If A and D are on the same arc, they are equal.

From the figure, A, B, C, D are on the circle with AC and BD intersecting inside. So A and D are on opposite sides relative to chord BC... Let me reconsider.

Using the correct approach:

∠ECD = ∠ACD = 20° (angle subtended by arc AD at C)
∠BAC = ∠DAC subtends arc DC...

Actually: ∠ABD and ∠ACD are in the same segment (subtend arc AD), so ∠ABD = ∠ACD = 20°.

In △ABE:
∠AEB = ∠BEC = 130° (vertically opposite... no, ∠AEB = 180° − 130° = 50°)

Wait: ∠AEB + ∠BEC = 180° (linear pair), so ∠AEB = 50°.

In △ABE:
$$\angle BAC + \angle ABE = 180° - \angle AEB$$
$$\angle BAC + \angle ABD = 180° - 50° = 130°$$

Hmm, but I need another relation.

Correct method:
∠ECD = 20° means ∠DCA = 20°, which is the angle subtended by chord AD at C.
So ∠DAB = ∠DCA = 20° ... no, ∠DAB subtends arc DB.

Let me use: In △BEC, ∠BEC = 130°.
Exterior angle of △BEC at E = ∠AEB = 50°.

In △AEB: ∠BAE + ∠ABE = 130° (exterior angle at E for the other triangle... )

Actually the cleanest approach:

∠ABD = ∠ACD = 20° (angles in same segment, subtending arc AD)

In △BEC:
∠CBE + ∠BCE = 180° − 130° = 50°
∠BCE = ∠BCA (part of it)

Using exterior angle theorem in △ECD:
∠BEC = ∠DCE + ∠CDE (exterior angle)
130° = 20° + ∠CDE
∠CDE = 110°

But ∠CDE = ∠CDB = 110°. Since ABCD is cyclic:
∠CAB + ∠CDB = 180° (opposite angles... no, they're not opposite angles of the quadrilateral necessarily).

Angles ∠BAC and ∠BDC subtend the same chord BC. If both A and D are on the major arc, they are equal. From the figure description, E is inside the circle, so A and D are on opposite arcs relative to BC. Therefore:
$$\angle BAC + \angle BDC = 180°$$
$$\angle BAC = 180° - 110° = 70°$$

$$\boxed{\angle BAC = 70°}$$

Given: ABCD is a cyclic quadrilateral. Diagonals AC and BD intersect at E. ∠DBC = 70°, ∠BAC = 30°.

Part 1: Find ∠BCD

∠DAC and ∠DBC are angles in the same segment (both subtend arc DC):
$$\angle DAC = ∠DBC = 70°$$

$$\angle DAB = \angle DAC + \angle CAB = 70° + 30° = 100°$$

Since ABCD is a cyclic quadrilateral, opposite angles are supplementary:
$$\angle DAB + \angle BCD = 180°$$
$$\angle BCD = 180° - 100° = 80°$$

$$\boxed{\angle BCD = 80°}$$

Part 2: Find ∠ECD, given AB = BC

Since AB = BC, triangle ABC is isosceles:
$$\angle BAC = \angle BCA = 30°$$

$$\angle BCD = 80°$$

$$\angle ECD = \angle BCD - \angle BCE = 80° - 30° = 50°$$

$$\boxed{\angle ECD = 50°}$$

Given: ABCD is a cyclic quadrilateral. Diagonals AC and BD are diameters of the circle.

To Prove: ABCD is a rectangle.

Proof:

Since AC is a diameter:
$$\angle ABC = 90° \quad \text{(angle in a semicircle)}$$
$$\angle ADC = 90° \quad \text{(angle in a semicircle)}$$

Since BD is a diameter:
$$\angle BAD = 90° \quad \text{(angle in a semicircle)}$$
$$\angle BCD = 90° \quad \text{(angle in a semicircle)}$$

Since all four angles of quadrilateral ABCD are 90°:
$$\angle A = \angle B = \angle C = \angle D = 90°$$

Also, since both diameters bisect each other (they pass through the centre), the diagonals bisect each other. A parallelogram with all right angles is a rectangle.

Hence, ABCD is a rectangle. Proved.

Given: ABCD is a trapezium with AB ∥ CD and AD = BC (non-parallel sides equal).

To Prove: ABCD is a cyclic quadrilateral.

Construction: Draw DE ⊥ AB and CF ⊥ AB.

Proof:

In △DAE and △CBF:
- DE = CF (perpendicular distance between parallel lines AB and CD is equal)
- AD = BC (given)
- ∠DEA = ∠CFB = 90°

By RHS congruence:
$$\triangle DAE \cong \triangle CBF$$

By CPCT:
$$\angle DAB = \angle CBA \quad \cdots (1)$$

Since AB ∥ CD:
$$\angle DAB + \angle CDA = 180° \quad \text{(co-interior angles)} \quad \cdots (2)$$

From (1): ∠CBA = ∠DAB

Substituting in (2):
$$\angle CBA + \angle CDA = 180°$$

Since the sum of one pair of opposite angles is 180°, ABCD is a cyclic quadrilateral.

Hence proved.

Given: Two circles intersect at B and C. Line ABD intersects the circles at A (first circle) and D (second circle). Line PBQ intersects the circles at P (first circle) and Q (second circle).

To Prove: ∠ACP = ∠QCD

Proof:

For the first circle (passing through A, B, P, C):
∠ACP and ∠ABP are in the same circle.

Since ABPC is a cyclic quadrilateral (A, B, P, C lie on the first circle):
$$\angle ACP = \angle ABP \quad \cdots (1)$$
(Angles subtended by arc AP at C and B — angles in the same segment)

For the second circle (passing through D, B, Q, C):
Since DBQC is a cyclic quadrilateral (D, B, Q, C lie on the second circle):
$$\angle QCD = \angle QBD \quad \cdots (2)$$
(Angles in the same segment, subtended by arc QD)

Now, ∠ABP and ∠QBD are vertically opposite angles:
$$\angle ABP = \angle QBD \quad \cdots (3)$$

From (1), (2) and (3):
$$\angle ACP = \angle QCD$$

Hence proved.

Given: △ABC. A circle is drawn with AB as diameter and another circle with AC as diameter. Let D be a point of intersection of these two circles (other than A).

To Prove: D lies on BC.

Proof:

Since AB is a diameter of the first circle and D lies on it:
$$\angle ADB = 90° \quad \text{(angle in a semicircle)} \quad \cdots (1)$$

Since AC is a diameter of the second circle and D lies on it:
$$\angle ADC = 90° \quad \text{(angle in a semicircle)} \quad \cdots (2)$$

Adding (1) and (2):
$$\angle ADB + \angle ADC = 90° + 90° = 180°$$

Since ∠ADB + ∠ADC = 180°, the rays DB and DC form a straight line.

Therefore, B, D and C are collinear, i.e., D lies on BC.

Hence proved.

Given: △ABC and △ADC are right triangles with common hypotenuse AC. ∠ABC = 90° and ∠ADC = 90°.

To Prove: ∠CAD = ∠CBD

Proof:

Since ∠ABC = 90°, point B lies on a circle with AC as diameter.

Since ∠ADC = 90°, point D also lies on a circle with AC as diameter.

Therefore, A, B, C and D are concyclic (all lie on the circle with AC as diameter).

Now, ∠CAD and ∠CBD both subtend the same chord CD in the same circle.

Since angles in the same segment are equal:
$$\angle CAD = \angle CBD$$

Hence proved.

Given: ABCD is a cyclic parallelogram.

To Prove: ABCD is a rectangle.

Proof:

Since ABCD is a cyclic quadrilateral, the sum of opposite angles is 180°:
$$\angle A + \angle C = 180° \quad \cdots (1)$$

Since ABCD is a parallelogram, opposite angles are equal:
$$\angle A = \angle C \quad \cdots (2)$$

From (1) and (2):
$$\angle A + \angle A = 180°$$
$$2\angle A = 180°$$
$$\angle A = 90°$$

Since ∠A = 90° and opposite angles of a parallelogram are equal:
$$\angle A = \angle C = 90°$$

Also, ∠B + ∠A = 180° (co-interior angles, since AB ∥ CD):
$$\angle B = 90°, \quad \angle D = 90°$$

Since all angles of the parallelogram are 90°, ABCD is a rectangle.

Hence proved.