We assume that \[ABCD\]be the given cyclic quadrilateral. \[PA\text{ }=\text{ }PD\text{ }\left[ Given \right]\] So, \[\angle PAD\text{ }=\angle PDA\text{ }\ldots \ldots \text{ }\left( 1 \right)\]...

Solution: Join \[AC,\text{ }PQ\text{ }and\text{ }BD.\] As \[ACQP\]is a cyclic quadrilateral \[\angle CAP\text{ }+\angle PQC\text{ }=\text{ }{{180}^{o}}~\ldots \ldots .\text{ }\left( i \right)\]...

Solution: Join \[AC\text{ }and\text{ }BD.\] Hence, we get \[\angle CAD\text{ }=\text{ }{{90}^{o}}~and\angle CBD\text{ }=\text{ }{{90}^{o}}\] [Angle is a semicircle is a right angle] And, \[AB\text{...

Solution: (i) Join \[RN\text{ }and\text{ }MS\] \[\angle RMS\text{ }=\text{ }{{90}^{o}}~\] [Angle in a semi-circle is a right angle] By angle sum property of \[\vartriangle RMS\] \[\angle RMS\text{...

Solution: To prove: \[\angle AOC\text{ }+\angle BOD\text{ }=\text{ }2\angle APC\] \[OA,\text{ }OB,\text{ }OC\text{ }and\text{ }OD\] are joined. \[AD\] is joined. Now, it’s seen that \[\angle...

Solution: (i) According to the given question, it’s clear that \[\angle ABC\text{ }=\text{ }{\scriptscriptstyle 1\!/\!{ }_2}\text{ }Reflex\text{ }(\angle COA)\] [Angle at the centre is double the...

Solution: According to the given question, Let \[O\text{ }and\text{ }O\]be the centres of two intersecting circles, where points of the intersection are \[P\text{ }and\text{ }Q\text{ }and\text{...

Solution: According to the given question, \[AB\text{ }=\text{ }AC\] So, \[\angle B\text{ }=\angle C\text{ }\ldots \text{ }\left( 1 \right)\] [Angles opposite to equal sides are equal] And,...

Solution: (i) Let’s expect that \[ABCD\]is a parallelogram which is inscribed in a circle. Hence, we know that \[\angle BAD\text{ }=\angle BCD\]  [Opposite angles of a parallelogram are equal] And...

Solution: Now, \[\angle BAD\text{ }+\angle BFE\text{ }=\text{ }{{96}^{o}}~+\text{ }{{84}^{o}}~=\text{ }{{180}^{o}}\] But these two are interior angles on the same side of a pair of lines \[AD\text{...

Solution: \[ABCD\]is a cyclic quadrilateral (i) Hence, \[\angle BAD\text{ }+\angle BCD\text{ }=\text{ }{{180}^{o}}\] [Pair of opposite angles in a cyclic quadrilateral are supplementary] \[\angle...

Solution: Join \[OB.\] Then,\[\angle OBA\text{ }=\text{ }{{90}^{o}}\] [Angle in a semi-circle is a right angle] Which means, \[OB\]is perpendicular to \[AE.\] Now, we know that the perpendicular...

Solution: (i) According to the given question, We know that \[\angle DCE\text{ }=\text{ }{{90}^{o}}~\angle CDE\] \[=\text{ }{{90}^{o}}-\text{ }{{40}^{o}}~=\text{ }{{50}^{o}}\] Hence, \[\angle...

Solution:  In quad. \[ABCD,\] \[\angle DAB\text{ }+\angle DCB\text{ }=\text{ }{{180}^{o}}\] \[{{27}^{o}}~+\angle CAB\text{ }+\text{ }{{83}^{o}}~=\text{ }{{180}^{o}}\] Hence, \[\angle CAB\text{...

Solution: (i) According to the given question, We know that \[\angle DBC\text{ }=\angle DAC\text{ }=\text{ }{{27}^{o}}\] [Angles subtended by the same chord on the circle are equal]   (ii)...

Solution: According to the given question, we get \[\angle ACB\text{ }=\text{ }{\scriptscriptstyle 1\!/\!{ }_2}\angle AOB\] \[=\text{ }{\scriptscriptstyle 1\!/\!{ }_2}\text{ }x\text{...

Solution: Join \[AD\] So, we get \[\angle ADC\text{ }=\text{ }{\scriptscriptstyle 1\!/\!{ }_2}\angle AOC\text{ }=\text{ }{\scriptscriptstyle 1\!/\!{ }_2}\text{ }x\text{ }{{110}^{o}}~=\text{...

Solution: According to the given question \[\angle ACB\text{ }=\text{ }{{90}^{o}}~\]  [Angle in a semi-circle is 90o] Also, \[\angle ABC\text{ }=\text{ }{{180}^{o}}-\angle ADC\] \[=\text{...

Solution: According to the given question, \[\angle CAB\text{ }=\text{ }{{75}^{o}}~and\angle CBA\text{ }=\text{ }{{50}^{o}}\] In \[\vartriangle ABC\], by angle sum property we have \[\angle...

Solution: (i) According to the given question, \[\angle ACB\text{ }and\angle ADB\]are in the same segment, So, \[\angle ADB\text{ }=\angle ACB\text{ }=\text{ }{{60}^{o}}\] (ii) Now, join \[OA\text{...

Solution: According to the given ques, \[\angle BCA\text{ }=\angle ADB\text{ }=\text{ }{{47}^{o}}\] [Angles subtended by the same chord on the circle are equal]

Solution: (i) According to the given question, By angle sum property of triangle ABD, \[\angle ADB\text{ }=\text{ }{{180}^{o}}-\text{ }{{75}^{o}}-\text{ }{{58}^{o}}~=\text{ }{{47}^{o}}\] Hence,...

Solution: According to the given question By angle sum property of a triangle we have \[\angle ACB\text{ }=\text{ }{{180}^{o}}-\text{ }{{49}^{o}}-\text{ }{{43}^{o}}~=\text{ }{{88}^{o}}\]

Solution: According to the given question We get, (i) \[\angle CDB\text{ }=\angle BAC\text{ }=\text{ }{{49}^{o}}\] (ii) \[\angle ABC\text{ }=\angle ADC\text{ }=\text{ }{{43}^{o}}\] [Angles subtended...

Solution: According to the given question, (i) In \[\vartriangle AOB,\]we have \[OA\text{ }=\text{ }OB\text{ }\left( radii \right)\] Thus, \[\angle OBA\text{ }=\angle OAB\] By angle sum property of...

Solution: According to the given question, Given, \[\angle AOB\text{ }=\text{ }{{140}^{o}}~\]and \[\angle OAC\text{ }=\text{ }{{50}^{o}}\] (i) So, \[\angle ACB\text{ }=\text{ }{\scriptscriptstyle...

Solution: The sum of angles in a quadrilateral is \[{{360}^{o}}\] Thus, \[\angle ADC\text{ }+\angle DAB\text{ }+\angle BCD\text{ }+\angle ABC\text{ }=\text{ }{{360}^{o}}\] \[{{75}^{o~}}+\text{...

Solution: According to the given question, it’s clear that In cyclic quadrilateral \[ABCD,\text{ }DC\text{ }||\text{ }AB\] And given, \[\angle DAB\text{ }=\text{ }{{105}^{o}}\] (i) Hence, \[\angle...

Solution: According to the given question, it’s clear that \[\angle DBA\text{ }=\angle CBA\text{ }=\text{ }{{90}^{o}}~\] [Angle in a semi-circle is a right angle] So, adding both \[\angle DBA\text{...

(i) (ii) SOLUTION: (i) According to the given question, \[\angle AOB\text{ }=\text{ }2\angle AOB\text{ }=\text{ }2\text{ }x\text{ }{{50}^{o}}~=\text{ }{{100}^{o}}\] [Angle at the center is double...

SOLUTION: (i) In the following figure, \[\angle BAD\text{ }=\text{ }{{90}^{o}}~\] [Angle in a semi-circle] Therefore, \[\angle BDA\text{ }=\text{ }{{90}^{o}}-\text{ }{{35}^{o}}~=\text{ }{{55}^{o}}\]...

(i) (ii) Solution: (i)In the following figure, \[b\text{ }=\text{ }{\scriptscriptstyle 1\!/\!{ }_2}\text{ }x\text{ }{{130}^{o}}\] [Angle at the center is double the angle at the circumference...

Solution: \[\angle AOB\text{ }=\text{ }2\angle ACB\] [Angle at the center is double the angle at the circumference subtend by the same chord] \[\angle ACB\text{ }=\text{ }{{70}^{o}}/\text{ }2\text{...

Solution: (i) In \[\vartriangle ABD,\] \[\angle DAB\text{ }+\angle ABD\text{ }+\angle ADB\text{ }=\text{ }{{180}^{o}}\] \[{{65}^{o}}~+\text{ }{{70}^{o}}~+\angle ADB\text{ }=\text{ }{{180}^{o}}\]...

Solution: Let’s join \[AC.\] And, let \[\angle OAC\text{ }=\angle OCA\text{ }=\text{ }x\] [Angles opposite to equal sides are equal] Thus, \[\angle AOC\text{ }=\text{ }{{180}^{o}}-\text{ }2x\] Also,...