Advertisements
Advertisements
प्रश्न
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).
उत्तर
\[\angle ADB = 90° \left( \text{ Angle in a semicircle } \right)\]
\[\angle ADC = 90° \left( \text{ Angle in a semicircle } \right)\]
\[\text{ So } , \angle ADB + \angle ADC = 90° + 90° = 180\]
\[\text{ Therefore, BDC is a line } . \]
\[\text{ Hence, the point of intersection of two circles lie on the third side } .\]
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.
Prove that the circle drawn with any side of a rhombus as diameter passes through the point of intersection of its diagonals.
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.
Prove that the perpendicular bisectors of the sides of a cyclic quadrilateral are concurrent.
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 quadrilateral inscribed in a circle with centre O. CD is produced to E such that ∠AED = 95° and ∠OBA = 30°. Find ∠OAC.
ABCD is a cyclic quadrilateral. M (arc ABC) = 230°. Find ∠ABC, ∠CDA, and ∠CBE.
In a cyclic quadrilaterals ABCD, ∠A = 4x, ∠C = 2x the value of x is
ABCD is a cyclic quadrilateral such that ∠A = 90°, ∠B = 70°, ∠C = 95° and ∠D = 105°.
