Question

prove that the perpendicular bisector of the sides of a cyclic quadrilateral are congruent

Answer 1

Answer:

Step-by-step explanation:

We have CEFD is a cyclic quadrilateral, let us take two points on the cyclic quadrilateral that is G and A. Draw a line segment between them which will become the chord of the circle. Join the line segments AC,AG and AD such that in ΔCGA and ΔDGA, we have

CA=DA (Radii of the circle)

AG=AG (common)

CG=DG ( G bisects CD)

Therefore, by SSS rule of congruency,

ΔCGA ≅ΔDGA.

Since, ∠AGC and ∠AGD are congruent and and are equal to 90° by construction, therefore we can conclude that the center A is on the perpendicular bisector of the drawn chord or any other chord which can be drawn. Thus, for any two chords, the perpendicular bisectors will be congruent at the center, hence perpendicular bisectors of the sides of a cyclic quadrilateral are congruent.