Solution: \[Arc\text{ }AC\text{ }subtends\angle AOC\]at the centre and \[\angle ADC\]at the remaining part of the circle. Thus, \[\angle AOC\text{ }=\text{ }2\angle ADC\] \[\angle AOC\text{ }=\text{...

Solution: Join \[EB\] Then, in cyclic quad. \[ABEP\] \[\angle APE\text{ }+\angle ABE\text{ }=\text{ }{{180}^{o}}~\ldots ..\text{ }\left( i \right)\][Opposite angles of a cyclic quad. are...

Solution: Given, \[AB\text{ }=\text{ }AD\text{ }=\text{ }DC\text{ }=\text{ }PB\text{ }and~\angle DBC\text{ }=\text{ }{{x}^{o}}\] Join \[AC\text{ }and\text{ }BD\] Proof: \[\angle DAC\text{ }=\angle...

Solution: Join \[DB\text{ }and\text{ }DC,\text{ }IB\text{ }and\text{ }IC\] Given, if \[\angle BAC\text{ }=\text{ }{{66}^{o~}}and~\angle ABC\text{ }=~{{80}^{o}}\] \[I\] is the incentre of the...

Solution: Join \[DB\text{ }and\text{ }DC,\text{ }IB\text{ }and\text{ }IC\] Given, if \[\angle BAC\text{ }=\text{ }{{66}^{o~}}and~\angle ABC\text{ }=~{{80}^{o}}\] \[I\] is the incentre of the...

Solution: Given, In cyclic quad. \[ABCD\] \[AF\text{ }||\text{ }CB\text{ }and\text{ }DA\] is produced to \[E\]such that \[\angle ADC\text{ }=\text{ }{{92}^{o}}~and\angle FAE\text{ }=\text{...

Given, \[\vartriangle ABC,\text{ }AB\text{ }=\text{ }AC\text{ }and\text{ }D\text{ }and\text{ }E\]are points on \[AB\text{ }and\text{ }AC\]such that \[AD\text{ }=\text{ }AE\] And, \[DE\] is joined....

Solution: In cyclic quadrilateral \[ABEC\] \[\angle BAC\text{ }+\angle BEC\text{ }=\text{ }{{180}^{o}}~\] [Opposite angles of a cyclic quadrilateral are supplementary] \[\angle BAC\text{ }+\text{...

Solution: (i) Given that \[BD\] is a diameter of the circle. And, the angle in a semicircle is a right angle. So, \[\angle BCD\text{ }=\text{ }90{}^\circ \] Also given that, \[\angle DBC\text{...

Let \[ABCD\] be a cyclic quadrilateral and \[PQRS\] be the quadrilateral formed by the angle bisectors of angle \[\angle A,\angle B,\angle C\text{ }and\angle D\] Required to prove: \[PQRS\] is a...

Solution: Join \[OE\] \[Arc\text{ }EC\] subtends \[\angle EOC\]at the centre and \[\angle EBC\]at the remaining part of the circle. \[\angle EOC\text{ }=\text{ }2\angle EBC\text{ }=\text{ }2\text{...

Suppose, \[\vartriangle ABC,\text{ }AB\text{ }=\text{ }AC\]and circle with \[AB\] as diameter is drawn which intersects the side \[BC\text{ }and\text{ }D\] And, join \[AD\] Proof: It’s seen that,...

Solution: Firstly, join \[OB\text{ }and\text{ }OC\] Proof: \[\angle BOC\text{ }=\text{ }2\angle BAC\text{ }=\text{ }2\text{ }x\text{ }{{30}^{o}}~=\text{ }{{60}^{o}}\] Now, in \[\vartriangle OBC\]...

Solution: Now, \[\angle ABD\text{ }=\angle ACD\text{ }=\text{ }{{30}^{o}}~\][Angles in the same segment] In \[\vartriangle ADB\] by angle sum property we have \[\angle BAD\text{ }+\angle ADB\text{...