Question

Topic: Circles
Class: 9th

$\sf \red{Question:}$
If two equal chords of a circle intersect within the circle, prove that the line joining the point of intersection to the centre makes equal angles with the chords.
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Answer 1

Step-by-step explanation:

Let AB and CD be the two equal chords. AB = CD.

Let the chords intersect at point E. Join OE.

Draw perpendiculars from the center O to the chords. The Perpendicular bisects the chord AB at M and CD at N.

To prove: ∠OEN = ∠OEM.

In ∆OME and ∆ONE,

∠M = ∠N = 90°

OE = OE

OM = ON (Equal chords are equidistant from the center.)

By RHS criteria, ∆OME and ∆ONE are congruent. So, by CPCT, ∠OEN = ∠OEM

Hence proved that the line joining the point of intersection of two equal chords to the center makes equal angles with the chords.

Answer 2

${\underline{\boxed{\pmb{\orange{\frak{ \; Given \; :- }}}}}}$

• AB = CD

$\\ \qquad{\rule{200pt}{2pt}}$

${\underline{\boxed{\pmb{\color{darkblue}{\frak{ \; To \; Prove \; :- }}}}}}$

• $\sf{ \angle OXA = \angle OXD }$

$\\ \qquad{\rule{200pt}{2pt}}$

${\underline{\boxed{\pmb{\pink{\frak{ \; ProoF \; :- }}}}}}$

${\underline{\pmb{\textbf{\textsf{ In \; ∆ \; OLE \; \& \; ∆ \; OME \; : }}}}}$

${\qquad \; \dashrightarrow \; \sf { OL = OM } \; \; \; \; \bigg( Equal \; Chords \; are \; Equidistant \bigg) }$

${\qquad \; \dashrightarrow \; \sf { OE = OE } \; \; \; \; \bigg( Common \bigg) }$

${\qquad \; \dashrightarrow \; \sf { \angle OLE = \angle OME } \; \; \; \; \bigg( Each \; 90° \bigg) }$

$\therefore \;$ ∆ OLE ≈ ∆ OME by RHS Congruence Criteria .

So :

$\implies \; \qquad {\underline{\boxed{\pmb{\purple{\sf{ \angle LEO = \angle MEO }}}}}} \; {\red{\bigstar}}$

$\therefore \;$ We have Proved that the line joining the point of intersection to the centre makes equal angles with the chords .

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