Question

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Q. Prove that the tangents drawn at the end points of a chord of a circle make equal angles with the chord.

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Answer 1

Step-by-step explanation:

Let AB be the chord of circle with centre M.

PA and PB are two tangents drawn from point P.

To prove: ∠PAM = ∠PBM

In ΔAPM & ΔBPM,

⇒ AP = BP {Tangents from an external point are equal}

⇒ ∠APM = ∠BPM

⇒ PM = PM {Common}

⇒ ΔPAM ≅ ΔPBM  {SAS congruence criterion}

⇒ ∠PAM = ∠PBM

Hope it helps!

Answer 2

$\bold{\underline{Figure \: \: is \: \: in \: \: the \: \: attached \: \: picture \:}}$.⬆⬆



$\red{\huge{ \underline{ \underline{ \overline{ \mid{ \bold{ \: \: SOLUTION \: \: { \mid}}}}}}}}$



$\bold{Let,} \\ \\ \bold{ PQ \: \: be \: \: the \: \: chord \: \: of \: \: the \: \: circle \: } \\ \bold{whose \: \: centre \: \: is \: \: \: O . \: }$

$\bold{Again} \\ \\ \bold{ AP \: \: and \: \: AQ \: \: be \: \: the \: \: tangents \: \: at \: \: P} \\ \\ \bold{ and \: \: Q \: \: from \: \: the \: \: common \: \: point \: \: A.}$

$\bold{We \: \: have \: \: to \: \: join \: \: AO \: \: and \: \:suppose \: } \\ \bold{ AO \: \: \: meets \: \: the \: \: chord \: \: at \: \: the \: \: point \: \: B.}$

$\bold{ \underline{To \: \: Prove \: : } \: \: \: \: \: \: \: \underline{ \angle{ APQ }= \angle{AQP}}}$



$\bold{ \underline{ \: PROOF \: \: : }}$



$\bold{In \: \: the \: \: ΔPAB \: \: and \: \: ΔQAB }\\ \\ \bold{PA =QA \: \: \: \: [Tangents \: \: from \: \: an } \\ \bold{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: external \: \: points \: \: to \: \: a \: } \\ \bold{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: circle \: \: are \: \: equal]} \\ \\ \bold{ \angle{PAB }= \angle{QAB} \: \: \: \: [AP \: \: and \: \: AQ \: \: are} \\ \bold{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: equally \: \: inclined \: \: to \: \:OA ]} \\ \\ \bold{AB = AB \: \: \: \: \: \: \: \: \: [Common \: \: side]} \\ \\ \bold{So,} \\ \\ \bold{ΔPAB ≅ΔQAB \: \: \: [S - A - S \: \: congruence]} \\ \\ \bold{Thus,} \\ \\ \bold{ \underline{ \: \: \angle{ APQ }= \angle{AQP} \: \: } \: \: \: \: \: \: \: \: \: [C.P.C.T.]}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \underline{ \underline{ \bold{ \: \: \: \: Hence \: \: Proved \: \: }}}$

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