# Circle

## Exercise 10.5 Part 2

Question 5: In the given figure; A. B, C and D are four points on a circle. AC and BD intersect at a point E such that ∠BEC = 130° and ∠ECD = 20°. Find ∠BAC.

**Answer:**In ΔEDC;

∠EDC + ∠ECD = 130°

Because external angle of a triangle is equal to sum of two opposite angles

Or, ∠EDC = 130° - 20° = 110°

Angles made on one side of a chord are equal

So, ∠BAC = ∠BDC = 110°

Question 6: ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If ∠DBC = 70°, ∠BAC is 30°, find ∠BCD. Further if AB = BC, find ∠ECD.

**Answer:** Given ABCD is a cyclic quadrilateral in which AB = BC, and ∠DBC = 70° and ∠BAC = 30°

∠DAC = ∠DBC = 70°

Because angles made one side of a chord are equal.

Now, ∠DAB = ∠DAC + ∠BAC

Or, ∠DAB = 70° + 30° = 100°

Opposite angles of cyclic quadrilateral are supplementary

Hence, ∠BCD = 180° - ∠DAB

Or, ∠BCD - 180° - 100° = 80°

In ΔABC we have AB = BC

So, ∠BAC = ∠BCA = 30°

So, ∠ECD = ∠BCD - ∠BCA

Or, ∠ECD = 80° - 30° = 50°

Question 7: If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle.

**Answer:** Given circle with centre O in which diagonals of cyclic quadrilateral ABCD intersect at O. This means AC = BD = diameter

To Prove: ABCD is a rectangle

In ΔAOB and ΔOBC;

AO = OC

BO = BO

All are radii

Hence, ΔAOB ≈ ΔOBC

So, ∠AOB = ∠BOC

As these angles make linear pair

So, ∠AOB + ∠BOC = 180°

Or, 2∠AOB = 180°

Or, ∠AOB = 90°

Now, in ΔAOB; as AO = OB

So, ∠OAB = ∠OBA

Or, ∠OAB + ∠OBA = 90°

Or, 2∠OAB = 90°

Or, ∠OAB = 45°

Similarly, it can be proved that ∠OBC = ∠OCB = ∠OCD = ∠ODC = 45°

This means that ∠OBA + ∠OBC = 90°

Similarly, it can be proven that all angles of quadrilateral ABCD are right angles

Hence, ABCD is a rectangle is proved

Question 8: If the non-parallel sides of a trapezium are equal, prove that it is cyclic.

**Answer:** Given trapezium ABCD in which AB||DC and AD = BC

Let us name the angles as 1, 2, 3, 4, 5, 6, 7 and 8; as shown in the figure.

In ΔADC and ΔBCD;

AD = BC (given)

DC = DC (common side)

Hence, ΔADC ≈ ΔBCD

So, ∠1 = ∠4 .............(1)

And ∠7 = ∠6 ……………(2)

In ΔDAB and ΔCBA

AD = BC (Given)

AB = AB (common side

Hence, ΔDAB ≈ ΔCBA

So, ∠8 = ∠5 …………..(3)

And ∠2 = ∠3 ……………(4)

Now, ∠1 + ∠4 + ∠7 + ∠6 + ∠8 + ∠5 + ∠2 + ∠3 = 360° (angle sum of quadrilateral)

From equations (1), (2), (3) and (4)

Or, ∠1 + ∠1 + ∠6 + ∠6 + ∠5 + ∠5 + ∠2 + ∠2 = 360°

Or, 2(∠1 + ∠2 + ∠6 + ∠5) = 360°

Or, ∠1 + ∠2 + ∠6 + ∠5 = 180°

Or, ∠DAB + ∠BCD = 180°

Similarly, following can be proven

∠ABC + ∠CDA = 180°

Here, angles of opposite vertices are supplementary

Hence, ABCD is a cyclic quadrilateral proved.