Advertisements
Advertisements
प्रश्न
ABCD is a cyclic quadrilateral in which BA and CD when produced meet in E and EA = ED. Prove that EB = EC.
Advertisements
उत्तर
EB = EC
Since, AD and BC are parallel to each other, so,
\[\angle ECB = \angle EDA \left( \text{ Corresponding angles } \right)\]
\[\angle EBC = \angle EAD \left( \text{ Corresponding angles} \right)\]
\[\text{ But } , \angle EDA = \angle EAD\]
\[\text{ Therefore } , \angle ECB = \angle EBC\]
\[ \Rightarrow EC = EB\]
\[ \text{ Therefore, } \bigtriangleup \text{ ECB is an isosceles triangle } .\]
APPEARS IN
संबंधित प्रश्न
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.
Prove that a cyclic parallelogram is a rectangle.
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.
Prove that the circles described on the four sides of a rhombus as diameters, pass through the point of intersection of its diagonals.
Prove that the perpendicular bisectors of the sides of a cyclic quadrilateral are concurrent.
In the given figure, O is the centre of the circle such that ∠AOC = 130°, then ∠ABC =
In a cyclic quadrilaterals ABCD, ∠A = 4x, ∠C = 2x the value of x is
In the figure, ▢ABCD is a cyclic quadrilateral. If m(arc ABC) = 230°, then find ∠ABC, ∠CDA, ∠CBE.
If a pair of opposite sides of a cyclic quadrilateral are equal, prove that its diagonals are also equal.
