Name: Proof of Angle in the Same Segment of a Circle
Uploaded: 2019-07-19
Duration: 1 min 26 s
Description: The most important and simple theorem from the chapter circle that angle in the same segment of the circle are equal. Let's derive this theorem.

Question

In the figure given below two equal chords $A B$ and $C D$ of a circle with centre $O$ intersect at right angles at $P .$ If $M$ and $N$ are mid-points of the chords $A B$ and $C D$ respectively, prove that $N O M P$ is a square.

Accepted Answer

Given: $A B = C D$
$∠ M E N = 90 °$
$M, N$ are mid point of $A B$ and $C D$

Prove: $D M E N$ is square

$OM ⊥ A B$ and $ON⊥CD$ (line passing through centre and bisecting the chord is perpendicular to chord )

In $∆ O N E$ and $△ OME$

$O E = O E$

$∠ O N E = ∠ O M E = (90 °)$

$O M = O N$ (When two chords are equal than they are equidistance from center)

$∆ O M E ≅ ∆ O N E$ (by SAS conruency)

$A B$

In $O M E N$

$O M = O N$

$M E = N E$

$∠ M E N = ∠ O M E = ∠ O N E = 90 °$

$∴ O M E N$ is square.

Hence, proved.

In the figure given below two equal chords $A B$ and $C D$ of a circle with centre $O$ intersect at right angles at $P .$ If $M$ and $N$ are mid-points of the chords $A B$ and $C D$ respectively, prove that $N O M P$ is a square.

Important Questions on Circle

Learn and Prepare for any exam you want !

In the figure given below two equal chords AB and CD of a circle with centre O intersect at right angles at P. If M and N are mid-points of the chords AB and CD respectively, prove that NOMP is a square.

Important Questions on Circle

Learn and Prepare for any exam you want !

In the figure given below two equal chords $A B$ and $C D$ of a circle with centre $O$ intersect at right angles at $P .$ If $M$ and $N$ are mid-points of the chords $A B$ and $C D$ respectively, prove that $N O M P$ is a square.