To do:We have to prove that the circle drawn with any side of a rhombus as diameter, passes through the point of intersection of its diagonals.Solution:Draw a circle with $Q$ as the centre.The circle passes through $A, B$ and $O$ as shown in the figure.This implies, $QA = QB = ... Read More

Given:ABCD is a parallelogram. The circle through \( \mathrm{A}, \mathrm{B} \) and \( \mathrm{C} \) intersect \( \mathrm{CD} \) (produced if necessary) at \( \mathrm{E} \).To do:We have to prove that \( \mathrm{AE}=\mathrm{AD} \).Solution:$ABCE$ is a cyclic quadrilateral. We know that, In a cyclic quadrilateral, the sum of the opposite ... Read More

Given:Two chords $AB$ and $CD$ of lengths $5\ cm$ and $11\ cm$ respectively of a circle are parallel to each other and are opposite side of its centre. The distance between $AB$ and $CD$ is $6\ cm$.To do:We have to find the radius of the circle.Solution:Let $r$ be the radius of ... Read More

Given:The length of two parallel chords of a circle are $6\ cm$ and $8\ cm$.The smaller chord is at a distance of $4\ cm$ from the centre.To do:We have to find the distance of the other chord from the centre.Solution:Let a circle with centre $O$ and two parallel chords $AB$ ... Read More

Given:Let the vertex of an angle \( \mathrm{ABC} \) be located outside a circle and let the sides of the angle intersect equal chords \( \mathrm{AD} \) and \( \mathrm{CE} \) with the circle.To do:We have to prove that \( \angle \mathrm{ABC} \) is equal to half the difference of ... Read More

To do:We have to prove that the line of centres of two intersecting circles subtends equal angles at the two points of intersection.Solution:Let two circles with centres $A$ and $A'$ intersect each other at $B$ and $B'$ respectively.In $\triangle BAA’$ and $\triangle B'AA’$$AB = AB'$          (Radii ... Read More

Given:\( \mathrm{ABC} \) and \( \mathrm{ADC} \) are two right triangles with common hypotenuse \( \mathrm{AC} \).To do:We have to prove that \( \angle \mathrm{CAD}=\angle \mathrm{CBD} \).Solution:We know that,Angles in the same segment are equal.This implies,$\angle CBD=\angle CAD$Hence proved.

Given:Circles are drawn taking two sides of a triangle as diameters.To do:We have to prove that the point of intersection of these circles lie on the third side.Solution:Let us draw a triangle $PQR$ and two circles having diameters as $PQ$ and $PR$ respectively.We know that, Angles in a semi-circle are ... Read More

Given:Two circles intersect at two points \( B \) and \( C \). Through \( \mathrm{B} \), two line segments \( \mathrm{ABD} \) and \( \mathrm{PBQ} \) are drawn to intersect the circles at \( A, D \) and \( P \), \( Q \) respectively.To do:We have to prove ... Read More

Given:The non-parallel sides of a trapezium are equal.To do:We have to prove that it is cyclic.Solution:Let $PQRS$ be a trapezium in which $PQ \| RS$ and $PS=QR$Draw $PM \perp RS$ and $QN \perp RS$In $\triangle PSM$ and $\triangle QRN$, $PS=QR$$\angle PMS=\angle QNR=90^o$$PM=QN$  (Perpendicular distance between two parallel lines is equal)Therefore, ... Read More