Advertisements
Advertisements
Question
If circles are drawn taking two sides of a triangle as diameters, prove that the point of intersection of these circles lie on the third side.
Advertisements
Solution
Consider a ΔABC.
Two circles are drawn while taking AB and AC as the diameter.
Let them intersect each other at D, and let D not lie on BC.
Join AD.
∠ADB = 90°...(Angle subtended by semi-circle)
∠ADC = 90° ...(Angle subtended by semi-circle)
∠BDC = ∠ADB + ∠ADC = 90° + 90° = 180°
Therefore, BDC is a straight line, and hence, our assumption was wrong.
Thus, point D lies on the third side BC of ΔABC.
RELATED QUESTIONS
ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If ∠DBC = 70°, ∠BAC is 30°, find ∠BCD. Further, if AB = BC, find ∠ECD.
If the non-parallel sides of a trapezium are equal, prove that it is cyclic.
Two circles intersect at two points B and C. Through B, two line segments ABD and PBQ are drawn to intersect the circles at A, D and P, Q respectively (see the given figure). Prove that ∠ACP = ∠QCD.
AC and BD are chords of a circle which bisect each other. Prove that (i) AC and BD are diameters; (ii) ABCD is a rectangle.
In any triangle ABC, if the angle bisector of ∠A and perpendicular bisector of BC intersect, prove that they intersect on the circumcircle of the triangle ABC.
Two chords AB and CD of lengths 5 cm 11cm respectively of a circle are parallel to each other and are on opposite sides of its centre. If the distance between AB and CD is 6 cm, find the radius of the circle.
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.
ABCD is a cyclic quadrilateral in ∠DBC = 80° and ∠BAC = 40°. Find ∠BCD.
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.
Prove that the perpendicular bisectors of the sides of a cyclic quadrilateral are concurrent.
ABCD is a cyclic quadrilateral in which BA and CD when produced meet in E and EA = ED. Prove that AD || BC .
In the given figure, O is the centre of the circle such that ∠AOC = 130°, then ∠ABC =
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.
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.
ABCD is a parallelogram. A circle through A, B is so drawn that it intersects AD at P and BC at Q. Prove that P, Q, C and D are concyclic.
