Advertisements
Advertisements
प्रश्न
ABCD is a cyclic quadrilateral in which BA and CD when produced meet in E and EA = ED. Prove that AD || BC .
Advertisements
उत्तर
If ABCD is a cyclic quadrilateral in which AB and CD when produced meet in E such that EA = ED, then we have to prove the following, AD || BC
It is given that EA = ED, so
Since, ABCD is cyclic quadrilateral
`x + angleABC = 180 ⇒ angleDAB = 180 - x`
And ; ` x + angleBCD = 180 ⇒ angle BCD = 180- x `
Now,
`angle DAB + angle ABC = x + 180 - x = 180`
Therefore, the adjacent angles `angleDAB ` and `angleABC` are supplementary
Hence, AD || BC
APPEARS IN
संबंधित प्रश्न
If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle.
Prove that the line of centres of two intersecting circles subtends equal angles at the two points of intersection.
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.
ABCD is a cyclic quadrilateral in BC || AD, ∠ADC = 110° and ∠BAC = 50°. Find ∠DAC.
If the two sides of a pair of opposite sides of a cyclic quadrilateral are equal, prove that its diagonals are equal.
Circles are described on the sides of a triangle as diameters. Prove that the circles on any two sides intersect each other on the third side (or third side produced).
In the given figure, ABCD is a cyclic quadrilateral in which AC and BD are its diagonals. If ∠DBC = 55° and ∠BAC = 45°, find ∠BCD.

ABCD is a cyclic quadrilateral. M (arc ABC) = 230°. Find ∠ABC, ∠CDA, and ∠CBE.

Find all the angles of the given cyclic quadrilateral ABCD in the figure.
