# Circle : Exercise 10.5 (Mathematics NCERT Class 9th)

Q. 1     In figure A, B and C are three points on a circle with centre O such that $\angle BOC$ = 30º and $\angle AOB$ = 60º. If D is a point on the circle other than the arc ABC, find $\angle ADC$.

Since are ABC makes $\angle AOC = \angle AOB + \angle BOC$ = 60º + 30º = 90º at the centre of the circle and $\angle ADC$ at a point on the remaining part of the circle.
Therefore $\angle ADC = {1 \over 2}\left( {\angle AOC} \right) = {1 \over 2} \times 90^o = 45^o$

Q. 2     A chord of a circle is equal to the radius of the circle. Find th eangle subtended by the chord at a point on the minor arc and also at a point on the major are.
Sol.

Let PQ be chord. Join OP and OQ.
It is given that PQ = OP = OQ (Since Chord = radius)
Therefore $\Delta$ OPQ is equilateral.
$\Rightarrow$ $\angle POQ$ = 60º Since are PBQ makes reflex POQ = 360º – 60º = 300º at centre of the circle and $\angle PBQ$ at a point in the minor arc of the circle.
Therefore $\angle PBQ = {1 \over 2}\left( {reflex\,\angle POQ} \right)$
$= {1 \over 2} \times 300^o = 150^o$
Similarly, $\angle PAQ = {1 \over 2}\left( {\angle POQ} \right) = {1 \over 2}$ (60º) = 30º
Hence, angle subtended by the chord on the minor are 150º and on the major chord = 30º.

Q.3      In figure $\angle PQR$ = 100º, there P, Q and R are points on a circle with centre O. Find $\angle OPR$.
Sol.

Since the angle subtended by an arc of a circle at its centre is twice the angle subtended by the same are at a point on the circumference.
Therefore Reflex $\angle POR = 2\angle PQR$
$\Rightarrow$ Reflex $\angle POR = 2 \times 100^\circ = 200^\circ$
$\Rightarrow$ $\angle POR$ = 360º – 200º = 160º
In $\Delta$ OPR, OP = OR [Radii of the same circle]
$\Rightarrow$ $\angle OPR = \angle ORP$
[Angles opp. to equal sides are equal]
and $\angle POR =$ 160º ......... (1) [Proved above]
In $\Delta OPR$,
$\angle OPR + \angle ORP + \angle PQR = 180^\circ$ [Angle sum property]
$\Rightarrow$ $160^\circ� + 2\angle OPR = 180^\circ$
$\Rightarrow$ $2\angle OPR = 180^\circ� - 160^\circ$
$\Rightarrow$ $2\angle OPR = 20^\circ$
$\Rightarrow$ $\angle OPR = 10^\circ$

Q.4      In figure $\angle ABC =$ 69º, $\angle ACB =$ 31º, Find $\angle BDC.$

In $\Delta$ ABC,
$\angle BAC + \angle ABC + \angle BCA = 180^o$
$\Rightarrow$ $\angle BAC + {69^o} + {31^o} = 180^\circ$
$\Rightarrow$ $\angle BAC = 180^o - \left( {69^o + 31^o} \right)$
$= 180^o - 100^o = 80^o$
Since angles in the same segment are equal
Therefore $\angle BDC = \angle BAC = 80^o$

Q.5      In figure A, B, C and D are four points on a circle. AC and BD intersect at a point E such that $\angle BEC$ = 130º and $\angle ECD$ = 20º.  Find $\angle BAC$. Sol.

$\angle CED + \angle CEB$ = 180º [Linear pair]
$\Rightarrow$ $\angle CED +$ 130º = 180º
$\Rightarrow$ $\angle CED =$ 180º – 130º
= 50º
In $\Delta$ ECD, $\angle EDC + \angle CED + \angle ECD$ = 180º
$\Rightarrow$ $\angle EDC +$ 50º + 20º = 180º
$\Rightarrow$ $\angle EDC =$ 180º – 50º – 20º
= 110º
$\Rightarrow$ $\angle BDC = \angle EDC$ = 110º
Since angles in the same segment are equal
Therefore $\angle BAC = \angle BDC =$ 110º.

Q.6     ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If $\angle DBC =$ 70º, $\angle BAC$ is 30º, find $\angle BCD$Further, if AB = BC, find $\angle ECD.$
Sol.

$\angle BDC = \angle BAC$ [Angles in the same segment]
$\Rightarrow$ $\angle BDC =$ 30º [Since $\angle BAC =$ 30º (given] In $\Delta$ BCD, we have
$\angle BDC + \angle DBC + \angle BCD$ = 180º [Sum of $\angle s$ of a $\Delta$]
$\Rightarrow$ 30º + 70º + $\angle BCD$ = 180º [Since $\angle DBC$ = 70º, $\angle BDC$ = 30º]
$\Rightarrow$ $\angle BCD$ = 180º – 30º – 70º = 80º
If AB = BC, than $\angle BCA = \angle BAC$ = 30º [Angles opp. to equal sides in a $\Delta$ are equal]
Now , $\angle ECD = \angle BCD - \angle BCE$
= 80º – 30º = 50º
[Since $\angle BCD =$ 80º (found above) and $\angle BCE = \angle BCA =$ 30º]
Hence, $\angle BCD =$ 80º and $\angle ECD =$ 50º

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

Diagonals AC and BD of a cyclic quadrilateral are diameters of the circle through the vertices A, B, C and D of the quad. ABCD.
To prove : Quadrilateral ABCD is a rectangle. Solution : - Since all the radii of the same circle are equal
Therefore OA = OB = OC = OD

$\Rightarrow$ $OA = OC = {1 \over 2}AC$
and $OB = OD = {1 \over 2}BD$
$\Rightarrow$ AC = BD
Therefore the diagonals of the quadrilateral ABCD are equal and bisect each other.
$\Rightarrow$ Quadrilateral ABCD is a rectangle.

Q.8     If the non- parallel sides of a trapezium are equal prove that it is cyclic.
Sol.

Given : Non - parallel sides AD and BC of a trapezium are equal.
To prove : ABCD is a cyclic trapezium.
Construction : Draw DE $\bot$ AB and CF $\bot$ AB. Proof : In order to prove that ABCD is a cyclic trapezium it is sufficient to prove that $\angle B + \angle D$ = 180º.
In $\Delta$s DEA and CFB, we have
$\angle DEA = \angle CFB$ [Each = 90º]
and DE = CF [Distance between two|| lines is always equal]
Therefore by RHS criterion of congruence, we have
$DEA \cong CFB$
$\Rightarrow$ $\angle A = \angle B\,\,and\,\,\angle ADE = \angle BCF$
(Corresponding parts of congruent triangles are equal)

Now, $\angle ADE = \angle BCF$
$\Rightarrow$ 90º + $\angle ADE =$ 90º + $\angle BCF$
$\Rightarrow$ $\angle EDC + \angle ADE = \angle FCD + \angle BCF$
[Since $\angle EDC =$ 90º and $\angle FCD =$ 90º]
$\Rightarrow$ $\angle ADC = \angle BCD$
$\Rightarrow$ $\angle D = \angle C$
Thus , $\angle A = \angle B\,\,and\,\,\angle C = \angle D$
Therefore , $\angle A + \angle B + \angle C + \angle D =$ 360º
[Since sum of the angles of a quad. is 360º]
$\Rightarrow$ $2\angle B + 2\angle D =$ 360º
$\Rightarrow$ $\angle B + \angle D =$ = 180º
Hence, ABCD is a cyclic trapezium.

Q.9     Two circles intersect at two points B and C. Through B, two line segments ABD and PBQ are drawn to intersect the circles at A, D and P, Q respectively (see figure). Prove that $\angle ACP = \angle QCD$

Since angles in the same segment are equal.
Therefore $\angle ACP = \angle ABP$ ... (1)
and $\angle QCD = \angle QBD$ ... (2)
Also $\angle ABP = \angle QBD$ ... (3) [Vertically opp. angles]
Therefore From (1), (2) and (3) , we have
$\angle ACP = \angle QCD$

Q.10    If circles are drawn taking two sides of a triangle as diameters, prove that the point of intersection of these circles lie on the third side.
Sol.

Given : Two circles are drawn with sides AB and AC of $\Delta$ ABC as diameters. The circle intersect at D. To prove : D lies on BC.
Construction : Join A and D.
Proof : Since AB and AC are the diameters of the two circles. [Given]
Therefore $\angle ADB =$ 90º [Angles in a semi- circle]
and, $\angle ADC =$ 90º [Angles in a semi-circle]
Adding we get $\angle ADB + \angle ADC$ = 90º + 90º = 180º
$\Rightarrow$ BDC is a straight line.
Hence, D lies on BC.

Q.11    ABC and ADC are two right triangles with common hypotenuse AC. Prove that $\angle CAD = \angle CBD$.
Sol.

$\Delta$ ABC and ADC are right $\angle d$ with common hypotenuse AC. Draw a circle with AC as diameter passing through B and D. Join BD.
Clearly, $\angle CAD = \angle CBD$ [Since Angles in the same segment are equal]

Q.12    Prove that a cyclic parallelogram is a rectangle.
Sol.

Given : ABCD is a parallelogram inscribed in circle.
To prove : ABCD is a rectangle.
Proof : Since ABCD is a cyclic parallelogram.
Therefore $\angle A + \angle C$ = 180º ... (1)
But $\angle A = \angle C$ ... (2)
From (1) and (2), we have
$\angle A = \angle C$ = 90º
Similarly, $\angle B = \angle D$ = 90º
Therefore Each angle of ABCD is of 90º
Hence, ABCD is a rectangle.

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