Mathematics
In the adjoining figure, ABC is an isosceles triangle in which AB = AC and circle passing through B and C intersects sides AB and AC at points D and E. Prove that DE || BC.
Answer
Given, AB = AC
∴ ∠ABC = ∠ACB (As angles opposite to equal sides are equal)
As BCED is a cyclic quadrilateral,
∠ADE = ∠BCE (∵ exterior angle of a cyclic quadrilateral is equal to the opposite interior angle.)
∴ ∠ADE = ∠ABC
Since these angles are equal and are corresponding this is the property of parallel lines,
∴ DE || BC.
Hence, proved that DE || BC.
Related Questions
Prove that a cyclic parallelogram is a rectangle.
In the figure (i) given below, ED and BC are two parallel chords of the circle and ABE, ACD are two st. lines. Prove that AED is an isosceles triangle.
Prove that a cyclic rhombus is a square.
In the figure (ii) given below, SP is the bisector of ∠RPT and PQRS is a cyclic quadrilateral. Prove that SQ = RS.
