Advertisements
Advertisements
प्रश्न
Prove that the circles described on the four sides of a rhombus as diameters, pass through the point of intersection of its diagonals.
Advertisements
उत्तर
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
संबंधित प्रश्न
ABC and ADC are two right triangles with common hypotenuse AC. Prove that ∠CAD = ∠CBD.
ABCD is a parallelogram. The circle through A, B and C intersect CD (produced if necessary) at E. Prove that AE = AD.
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.
Bisectors of angles A, B and C of a triangle ABC intersect its circumcircle at D, E and F respectively. Prove that the angles of the triangle DEF are 90°-A, 90° − `1/2 A, 90° − 1/2 B, 90° − 1/2` C.
In the figure, `square`ABCD is a cyclic quadrilateral. Seg AB is a diameter. If ∠ ADC = 120˚, complete the following activity to find measure of ∠ BAC.
`square` ABCD is a cyclic quadrilateral.
∴ ∠ ADC + ∠ ABC = 180°
∴ 120˚ + ∠ ABC = 180°
∴ ∠ ABC = ______
But ∠ ACB = ______ .......(angle in semicircle)
In Δ ABC,
∠ BAC + ∠ ACB + ∠ ABC = 180°
∴ ∠BAC + ______ = 180°
∴ ∠ BAC = ______
Prove that the centre of the circle circumscribing the cyclic rectangle ABCD is the point of intersection of its diagonals.
In the given figure, ABCD is a cyclic quadrilateral in which ∠BAD = 75°, ∠ABD = 58° and ∠ADC = 77°, AC and BD intersect at P. Then, find ∠DPC.
ABCD is a cyclic quadrilateral such that AB is a diameter of the circle circumscribing it and ∠ADC = 140º, then ∠BAC is equal to ______.
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.
