Question

Prove that "If a line touches a circle and from the point of contact a chord is drawn, the angle
which this chord makes with the given line are equal respectively to the angles formed in the
corresponding alternate segments."

Answer 1

Given that : AB and CD are two equal chords. And, M, N are mid point of chord AB and CD respectively.

To prove : ∠AMN=∠CNM and ∠BMN=∠DNM

Construction : Join OM and ON

Proof :

Since the line segment joining the centre of a circle to the midpoint of a chord is perpendicular to the chord.

Since AB and CD are equal chords, they are equidistant from the other.

i.e., OM =ON

In ΔOMN,

OM=ON (Proved)

∠OMN=∠ONM (Angles opposite to equal sides) .....1

∠OMA=∠ONC (each 90°) .....2

∠OMB=∠OND (each 90°) .....3

Subtracting 2 from 1, we have,

∠OMA-∠OMN=∠ONC-∠ONM

⇒∠AMN=∠CNM

Adding 1 and 3, we have,

∠OMB+∠OMN=∠OND+∠ONM

⇒∠BMN=∠DNM

Hence Proved.

❤️BRAINLIEST PLS❤️