Question

N
17. A diameter PQ bisects two chords AB and CD at M and N respectively. Prove that AB II CD.​

Answer 1

$\huge\mathcal\colorbox{white}{{\color{magenta}{『ANSWER』}}}$

$\bf\bold\blue{⇝}$ A diameter PQ bisects two chords AB & CD at M & N respectively.

$\displaystyle\color{blue}{TO FIND :}$

$\green{↬}$ 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,

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

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

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

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

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

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

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

$➜ \rm\angle OMA = \angle OMB = 90°$

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

$\rm\angle OMA = \angle NMA$

$\rm\angle OND = \angle MND$

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

$➜\Large\bf\red{\:{AB}\:\|\:{CD}\:}$

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