Question

36. In the following figure O is the center of a circle. If AB and AC are chords of the circle.
such that AB = AC, OP is perpendicular to AB and OQ is perpendicular to AC then prove that PB = QC
pls tell guys ​

Answer 1

Given: AB and AC are two chords of the circle. O is the center of a circle.

• AB = AC
• OP $\perp$ AB
• OQ$\perp$ AC

To Prove: PB = QC

⠀⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀⠀⠀⠀⠀⠀⠀⠀

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AnswEr :

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$\underline{\bf{\dag} \:\mathfrak{As\;we\;know\: that\: :}}$⠀⠀⠀⠀

• Perpendicular from the center of a circle bisects the chord.

Therefore,

$:\implies\sf AM = MB \qquad\quad\bigg\lgroup\bf Equation\;(I)\bigg\rgroup$

Similarly,

$:\implies\sf AN = NC \qquad\quad\bigg\lgroup\bf Equation\;(II)\bigg\rgroup$

⠀⠀$\underline{\bf{\dag} \:\mathfrak{From \; both \; Equations\: :}}$⠀⠀⠀⠀

$:\implies\sf MB = NC$

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Given that,

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$:\implies\sf AB = AC \\\\\\:\implies\sf \dfrac{AB}{2} = \dfrac{AC}{2} \qquad\quad\bigg\lgroup\bf Dividing\;by\;(2)\bigg\rgroup$

Now,

• OP and OQ are the two radii of the circle.

Therefore, OP = OQ.

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Also,

• Equal chords are equidistant from the center of the circle.

Therefore, OM = ON.

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$:\implies\sf OP = OQ \; and\; OM = ON \qquad\quad\bigg\lgroup\bf Subtracting\; the \; Equations\; \bigg\rgroup \\\\\\:\implies\sf OP - OM = OQ - ON \\\\\\:\implies\sf PM = QN$

⠀⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀⠀⠀⠀⠀⠀⠀⠀

Now, In ∆MPB and ∆NQC

$:\implies\sf \angle PMB = \angle QNC \qquad\quad\bigg\lgroup\bf Each\; 90^{\circ} \bigg\rgroup$

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Now,

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By SAS (Side Angle Side) congruence rule, If two sides and the angle of one triangle are equal to two sides and the angle of second triangle, then the triangles are congruent.

Therefore,

• ∆MPB ≅ ∆NQC

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By CPCT (Corresponding parts of congruence Triangles),

• PB = QC.

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$\qquad\quad\dag\:{\underline{\underline{\sf{\purple{Hence\; Proved!!}}}}}$

Answer 2

Given: AB and AC are two equal chords of a circle with centre O.

OP⊥AB and OQ⊥AC.

To prove : PB = QC

Proof: OP⊥AB

⇒AM = MB .. (⊥ from centre bisects the chord) →(i)

Similarly, AN = NC →(ii)

But, AB = AC

⇒ AB/2 = AC/2

⇒MB = NC →(iii) [ From (i) and (ii) ]

Also, OP = OQ (Radii of the circle)

and OM = ON (Equal chords are equidistant from the centre)

⇒OP − OM = OQ−ON

⇒MP = NQ ....(iv) (From figure)

In ΔMPB and ΔNQC, we have

∠PMB =∠QNC (Each =90°)

MB = NC ( From (iii) )

MP = NQ ( From (iv) )

∴ΔPMB ≅ ΔQNC (SAS)

⇒PB = QC (CPCT)