Question

Class 9

Exercise- 10.4

No. 3

Question-
$\sf \orange{If \: two \: equal \: c hords \: of }\\ \sf \orange{ a \: circle \: intersect \: within \: }\\ \sf \orange{the \: circle, \: prove \: that \: }\\ \sf \orange{the \: line \: joining \: the \: point \: } \\ \sf \orange{of \: intersection \: to \: the \: centre }\\ \sf \orange{ \: makes \: equal \: angles \: with \: the \: chords.}\ \textless \ br /\ \textgreater$

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Answer 1

Naming:

Let the center of the circle be O.

And let the two chords of equal length be AB and CD.

Let the point of intersection of AB and CD be H.

[Refer to the attachment]

To prove:

Line joining the point of intersection of the two chords and the center of the circle makes equal angles with AB and CD.

i.e, ∠OHE = ∠OHF

Construction:

Draw a perpendicular from O to AB and another one from O to CD. Name them OE and OF respectively.

Proof:

In ΔOHE & ΔOHF;

➝ OH = OH (common side)

➝ ∠OEH = ∠OFH = 90° (OE ⊥ AB ; OF ⊥ CD)

➝ OE = OF (Equal chords are equidistant from the center of the circle)

∴ ΔOHE ≅ ΔOHF. (RHS Congruence Criterion)

Using CPCT (Corresponding parts of congruent triangles are equal) we can say that:

➝ ∠OHE = ∠OHF

Which is the same as;

➝ ∠OHB = ∠OHD

Hence proved.

Answer 2

Answer:

Given :-

• AB and CD are chord intersecting at E.
• AB = CD
• PQ is a diameter.

To Prove :-

• $\angle$BEQ = $\angle$CEQ.

Construction :-

• OM $\perp$ AB.
• ON $\perp$ CD.
• OE is joined.

Proof :-

$\mapsto$ In ∆OEM and ∆OEN.

Then, OM = ON

And, OE = OE (Common)

$\implies$ $\angle$ OME = $\angle$ ONE.

$\implies$ ∆OEM ≊ ∆OEN [RHS Congruence Rule]

Thus,

$\implies$ $\angle$MEO = $\angle$NEO [CPCT]

$\implies$ $\angle$BEQ = $\angle$CEQ

$\longmapsto$ Hence, Proved.