Question

M and N are the mid-points of two equal chords AB and CD respectively of a circle with centre O. then prove that angle BMN is equal to DNM.

Answer 1

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★Refer to the attachment.

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Answer 2

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M and N are the mid-points of two equal chords AB and CD respectively of a circle with centre O. then prove that angle BMN is equal to DNM.

GIVEN:

M and N are the mid-points of two equal chords.

AB and CD respectively of a circle with centre O.

R.T.P:

Angle BMN is equal to DNM

SOLUTION:

$\: OM \: bisects \: AB \: and \:ON \: bisects \: CD \\ \therefore Drop \: OM \perp AB \: and \: ON \perp CD \ \:$

(Perpendicular drawn from the centre of a circle to a chord bisects it)

$BM = \frac {1}{2}AB =\frac{1}{2}CD= DN-------------1$

By applying Pythagoras theorem:

$</p><p>OM²=OB²-MB² \\ = OD²-DN² \\=ON {}^{2}$

$\therefore \: OM=ON$

(Angles opposite to equal sides are equal)

$\angle\: OMB=\angle \: OND = 90 \degree$

Subtracting (2) from above, we get:

$\angle \: BMN = \angle DNM$

Hence proved

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