Question

17. A diameter PQ bisects two chords AB and CD at Mand N respectively. Prove that AB || CD.

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Answer 1

Given : A diameter PQ bisects two chords AB and CD at M and N respectively.

To Find : Prove that AB || CD.

Solution:

Diameter of circle always pa ss through center

diameter PQ pa ssed through center (let say O)

ΔOAM and ΔOBM

OA = OB  radius

AM = BM    ( as chord is bisected)

OM = OM  common

ΔOAM ≅ ΔOBM  (SSS)

=> ∠OMA = ∠OMB  CPCT

∠OMA +  ∠OMB = 180°   ( Linear Pair)

=> ∠OMA = ∠OMB  = 90°

Similarly    ∠ONC = ∠OND  = 90°

∠OMA = ∠NMA

∠OND  = ∠MND

∠NMA  = ∠MND  = 90°   ( alternate interior angles)

=> AB || CD

QED

Hence proved

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Answer 2

$\huge\mathcal\colorbox{red}{{\color{white}{ANSWER࿐}}}$

$\huge\color{navy}{\textbf{\textsf{ANSWER :}}}$

$\red\bigstar$ A diameter PQ bisects two chords AB & CD at M & N respectively.

$\huge\color{navy}{\textbf{\textsf{TO FIND :}}}$

$\red\bigstar$ Prove that AB || CD.

$\huge\color{navy}{\textbf{\textsf{SOLUTION :}}}$

Diameter of circle always passes through the center,

$\red\bigstar$ Let the center be O.

Diameter PQ passed through the center O,

$\Large\bf\red{In\:\triangle{OAM}\:\&\:\triangle{OBM}\:}$

$\Large \rm{ { {OA = OB \: (Radius) }}}$

$\Large \rm{ { {AM = BM \: (Cord \: is \: bisected) }}}$

$\Large \rm{ { {OM = OM \: (Common) }}}$

$\Large \rm\triangle \: OAM ≅ \triangle \: OBM \: (SSS)$

$➜\Large \rm\angle OMA = \angle OMB \: (CPCT)$

$\Large \rm\angle OMA + \angle OMB = 180° \: (Linear \: pair)$

$➜\Large \rm\angle OMA = \angle OMB = 90°$

$\Large \rm{ {{Similarly }}}\Large \rm\angle ONC = \angle OND = 90°$

$\Large \rm\angle OMA = \angle NMA$

$\Large \rm\angle OND = \angle MND$

$\Large \rm\angle NMA = \angle MND = 90° \: (Alternate \: interior \: angle)$

$➜\Large\bf\red{\:{AB}\:\|\:{CD}\:}$

$\huge\boxed{{\sf\red{Hence \: proved}}}$

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