Advertisements
Advertisements
Question
Prove that the circles described on the four sides of a rhombus as diameters, pass through the point of intersection of its diagonals.
Advertisements
Solution
Here, ABCD is a rhombus; we have to prove the four circles described on the four sides of any rhombusABCD pass through the point of intersection of its diagonals AC and BD.
Let the diagonals AC and BD intersect at O.
We know that the diagonals of a rhombus intersect at right angle.
Therefore,
`angle AOB = `90°
`angleBOC = `90°
`angle COD ` = 90°
`angle AOD ` = 90°
Now, `angle AOB ` = 90 means that circle described on AB as diameter passes through O.
Similarly the remaining three circles with BC, CD and AD as their diameter will also pass through O.
Hence, all the circles with described on the four sides of any rhombus ABCD pass through the point of intersection of its diagonals AC and BD.
APPEARS IN
RELATED QUESTIONS
Prove that ‘Opposite angles of a cyclic quadrilateral are supplementary’.
A chord of a circle is equal to the radius of the circle. Find the angle subtended by the chord at a point on the minor arc and also at a point on the major arc.
Prove that the line of centres of two intersecting circles subtends equal angles at the two points of intersection.
In the given figure, ABCD is a cyclic quadrilateral. Find the value of x.
ABCD is a cyclic quadrilateral in ∠BCD = 100° and ∠ABD = 70° find ∠ADB.
PQRS is a cyclic quadrilateral such that PR is a diameter of the circle. If ∠QPR = 67° and ∠SPR = 72°, then ∠QRS =
In the given figure, O is the centre of the circle such that ∠AOC = 130°, then ∠ABC =
ABCD is a cyclic quadrilateral such that ∠A = 90°, ∠B = 70°, ∠C = 95° and ∠D = 105°.
If a pair of opposite sides of a cyclic quadrilateral are equal, prove that its diagonals are also equal.
If non-parallel sides of a trapezium are equal, prove that it is cyclic.
