Question

If two equal chords of a circle intersect within the circle, prove that the line joining the point of intersection to the centre makes equal angles with the chords.​

Answer 1

Given:

Two equal chords AB and CD of a circle with centre O intersecting at point E within the circle.

To Prove:

∠OEA = ∠OED

Construction:

OM⊥AB, ON⊥CD

Proof:

Consider ΔOME and ΔONE

OE = OE [Common side]

OM = ON [Equal chords are equidistant from the centre]

∠OME = ∠ONE = 90° [OM⊥AB, ON⊥CD]

ΔOME ≅ ΔONE

∠OEA = ∠OED [Corresponding angles]

Hence, the line joining the points of intersection of two equal chords to the center makes equal angles with the chords.

Answer 2

Answer:

Step-by-step explanation:

❣Holla user❣

Here is ur ans⬇️⬇️⬇️

Given:-

AB and DC are the chords

M is the middle point of both the chords

Let AB and CD are the two equal chords of a circle having center O

Again let AB and CD intersect each other at a point M.

Now, draw OP perpendicular AB and OQ perpendicular CD

From the figure,

In ΔOPM and ΔOQM,

OP = OQ {equal chords are equally distant from the cntre}

∠OPM = ∠OQM

OM = OM {common}

By SAS congruence criterion,

ΔOPM ≅ ΔOQM

So, ∠OMA = ∠OMD

or ∠OMP = ∠OMQ {by CPCT}

Thus, the line joining the point of intersection to the center makes equal angles with the chords.

Referred the figure given below⬇️⬇️

Hope it helps u^_^