Question

Ab and CD are two equal chords of a circle with Centre O intersect each other at point as CM perpendicular a b and O and perpendicular to CD state in the following statement are true I am is equal to triangle o m s is congruent to triangle o n s

Answer 1

Answer:

Chords of the circle = AB and CD (Given)

Centre of the circle = O (Given)

Intersection point at right angle of the circle = P.

Since all the angles of OMPN are 90°, thus

OM ⊥ AB and ON ⊥ CD

in quadrilateral OMPN

∠OMP = ∠ONP =∠MPN = 90°(given)

= ∠MON = 90°

Therefore OMPN is a rectangle --- eq 1

Since, the perpendicular distance of equal chords from the centre of the circle are always equal, hence

= OM = ON --- eq 2

From equation (1) and (2) it can be concluded that the adjacent sides of a rectangle are equal and thus all sides are equal

Hence OMPN is a square.

refer attachment for figure.