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(i) 30°, 75°, 75°

(ii) 40°, 70°, 70°

(iii) 20°, 80°, 80°

(iv) 100°, 40°, 40°

The other two angles are equal. So they are 45°, 45°

AB = AD

∴ ∆ ABD is an isosceles triangle.

∴ ∠ABD = ∠ADB

It is given that ∠ABC = ∠ADC

∴ ∠CBD = ∠CDB

∴ CD = CB

∴ ∆ BCD is an isosceles triangle.

(a) OA = OB = OC, AB = BC

∆ OAB and ∆ OBC are equal triangles.

∴ ∠AOB and ∠BOC are equal which are opposite to the equal sides AB and BC.

(b) If OA = AB then ∆ OAB is an equilateral triangle.

If OB = BC, ∆ OBC is equilateral triangle. 

∴ ∠AOB = ∠BOC = 60°

(c) Each angle at O is 60°. The angle at the centimeter is O is 360° and 6 triangles can be drawn.

OA = OB

∴ ∆ OAB is an isosceles triangle.

∴ ∠A = ∠B

When ∆ OMA and ∆ OMB are considered, OM is the common side

∠AMO = ∠BMO = 90 ∠AOM = ∠BOM

One side of the triangle and angles at the ends of sides are equal. So the other two sides are also equal.

∴ AM = MB

∴ M is the midpoint of AB.

Given AB = 12 cm and M is the mid-point of AB.

∴ AM = MB = 6 cm

In quadrilateral AMCD,

AM = CD

AB||CD ∴ AM||CD

∴ AMCD is a parallelogram.

∴ ∠AMD = ∠CDM (Alternate interior angles)

∠ADM = ∠CMD (Alternate interior angles)

∠A = ∠DCM = 40° = ∠CMB

∴ ∠MCB = 80° [180 – (60 + 40)]

(i) The angles of ∆AMD, ∆MBC and ∆DCM are 40°, 60° and 80° respectively.

(ii) Both of them are parallelograms.

∆ OAB is an isosceles triangle.

∴ ∠A =∠B

∠O = 60

∴ ∠OAB is an equilateral triangle so perimeter = 3 + 3 + 3 = 9 cm.

OA = OB (radius of circle)

∆AOB is a isosceles triangle; ∴ ∠A = ∠B ∠O = 60°

∴ ∠A = ∠B = 60°

AC = BD

AB is the common side.

The angles between the sides AC, AB and BD, AB are equal.

∴ BC = AD

The opposite sides of quadrilateral ∆CBD are equal. The angles opposite to equal sides of triangles ∆ABC and ∆ABD are equal. So the opposite angles in quadrilateral ACBD are also equal.

∴ ACBD is a parallelogram.

When the diagonal of a quadrilateral with equal opposite sides is drawn, we get two equal triangles. The angels opposite to the diagonal in the triangles are equal. That is the opposite sides and angles in the quadrilateral are equal. So the quadrilateral is a parallelogram.

DC = AB ………. (1)

AD = CB

QC = AP ……. (2) (as ABCD is a parallelogram)

(1) – (2) ⇒ DC – QC = AB – AP; 

∴ DQ = PB When, ∆ APD, ∆ CQB are considered.

AD = CB

AP = QC

∠A = ∠C, The two sides and angle formed by them in these triangles are equal. So the third sides PD and BQ are equal.

∴ Two pairs of opposite sides in the quadrilateral PBQD are equal. So PBQD is a parallelogram.

The sides of ∆ DEF and ∆ ABC are equal. The angles opposite to equal sides are equal.

∠E = ∠B, ∠F = ∠C, ∠D = ∠A

But ∠C = 180 – (∠B + ∠A)

∠P = 180 – (∠B + ∠D)

∴ ∠D = ∠A

∴ ∠C = ∠P = 180 – (∠B + ∠A)

∴ ∠C = ∠P

C = 80° (Use the property that the sum of three angles of a triangle is 180°)

AB = QR 

∴ ∠C = ∠P 

∠C = 80° 

∴ ∠P = 80°

BC = RP 

∴ ∠A = ∠Q 

∠A = 40° 

∴ ∠Q = 40°

CA = PQ 

∴ ∠B = ∠R 

∠B = 60° 

∴ ∠R = 60° 

(The angle opposite to equal sides are equal)