## In any triangle ABC, if the angle bisector of ∠A and perpendicular bisector of BC intersect,

[question] Question. In any triangle ABC, if the angle bisector of ∠A and perpendicular bisector of BC intersect, prove that they intersect on the circum circle of the triangle ABC. [/question] [solution] Solution: Let perpendicular bisector of side BC and angle bisector of ∠A meet at point D. Let the perpendicular bisector of side BC intersect it at E. Perpendicular bisector of side BC will pass through circumcentre O of the circle. ∠BOC and ∠BAC are the angles subtended by arc BC at the centre...

## Two congruent circles intersect each other at points A and B.

[question] Question. Two congruent circles intersect each other at points A and B. Through A any line segment PAQ is drawn so that P, Q lie on the two circles. Prove that BP = BQ. [/question] [solution] Solution: $\mathrm{AB}$ is the common chord in both the congruent circles. $\therefore \angle \mathrm{APB}=\angle \mathrm{AQB}$ In $\triangle B P Q$, $\angle \mathrm{APB}=\angle \mathrm{AQB}$ $\therefore B Q=B P$ (Angles opposite to equal sides of a triangle) [solution]...

## Two chords AB and CD of lengths 5 cm 11cm respectively of a circle are parallel to each other and are on opposite sides of its centre.

[question] Question. Two chords AB and CD of lengths 5 cm 11cm respectively of a circle are parallel to each other and are on opposite sides of its centre. If the distance between AB and CD is 6 cm, find the radius of the circle. [/question] [solution] Solution: Draw $O M \perp A B$ and $O N \perp C D$. Join $O B$ and $O D$. $\mathrm{BM}=\frac{\mathrm{AB}}{2}=\frac{5}{2}$ (Perpendicular from the centre bisects the chord) $\mathrm{ND}=\frac{\mathrm{CD}}{2}=\frac{11}{2}$ Let $O N$ be $x$. Therefor...

## Prove that line of centres of two intersecting circles subtends

[question] Question. Prove that line of centres of two intersecting circles subtends equal angles at the two points of intersection. [/question] [solution] Solution: Let two circles having their centres as $O$ and $O^{\prime}$ intersect each other at point $A$ and $B$ respectively. Let us join $O O^{\prime}$. In $\triangle \mathrm{AO} \mathrm{O}^{\prime}$ and $\mathrm{BO} \mathrm{O}^{\prime}$, OA $=$ OB (Radius of circle 1) $\mathrm{O}^{\prime} \mathrm{A}=\mathrm{O}^{\prime} \mathrm{B}$ (Radius ...

## If circles are drawn taking two sides of a triangle as diameters,

[question] Question. 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. [/question] [solution] Solution: Consider a $\triangle \mathrm{ABC}$. Two circles are drawn while taking $A B$ and $A C$ as the diameter. Let they intersect each other at $D$ and let $D$ not lie on $B C$. Join AD. $\angle A D B=90^{\circ}$ (Angle subtended by semi-circle) $\angle \mathrm{ADC}=90^{\circ}$ (Angle subtended by semi-circl...

## In the given figure, ∠ABC = 69°,

[question] Question. In the given figure, $\angle A B C=69^{\circ}, \angle A C B=31^{\circ}$, find $\angle B D C$. [/question] [solution] Solution: In $\triangle A B C$ $\angle B A C+\angle A B C+\angle A C B=180^{\circ}$ (Angle sum property of a triangle) $\Rightarrow \angle B A C+69^{\circ}+31^{\circ}=180^{\circ}$ $\Rightarrow \angle B A C=180^{\circ}-100^{\circ}$ $\Rightarrow \angle B A C=80^{\circ}$ $\angle B D C=\angle B A C=80^{\circ}$ (Angles in the same segment of a circle are equal) [/s...