Advertisements
Advertisements
Question
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).
Advertisements
Solution

\[\angle ADB = 90° \left( \text{ Angle in a semicircle } \right)\]
\[\angle ADC = 90° \left( \text{ Angle in a semicircle } \right)\]
\[\text{ So } , \angle ADB + \angle ADC = 90° + 90° = 180\]
\[\text{ Therefore, BDC is a line } . \]
\[\text{ Hence, the point of intersection of two circles lie on the third side } .\]
APPEARS IN
RELATED QUESTIONS
If the non-parallel sides of a trapezium are equal, prove that it is cyclic.
Prove that a cyclic parallelogram is a rectangle.
In a cyclic quadrilateral ABCD, if ∠A − ∠C = 60°, prove that the smaller of two is 60°
ABCD is a cyclic quadrilateral in ∠DBC = 80° and ∠BAC = 40°. Find ∠BCD.
ABCD is a cyclic quadrilateral in ∠BCD = 100° and ∠ABD = 70° find ∠ADB.
Prove that the circles described on the four sides of a rhombus as diameters, pass through the point of intersection of its diagonals.
If the two sides of a pair of opposite sides of a cyclic quadrilateral are equal, prove that its diagonals are equal.
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.
ABCD is a cyclic quadrilateral such that AB is a diameter of the circle circumscribing it and ∠ADC = 140º, then ∠BAC is equal to ______.
