Question
If two intersecting chords of a circle make equal angles with the diameter passing through their point of intersection, prove that the chords are equal.
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Answer
Given: $A B$ and $C D$ are two chords of a circle with centre $O$, intersecting at point $E . P Q$ is a diameter through $E$, such that $\angle AEQ =\angle DEQ$.
To prove: $A B=C D$
Construction: Draw $OL \perp AB$ and $OM \perp CD$
Proof: $\angle LOE +\angle LEO +\angle OLE =180^{\circ}$ (Angle sum property of a triangle)
$\Rightarrow \angle LOE+\angle LEO+90^{\circ}=180^{\circ}$
$\angle LOE +\angle LEO =90^{\circ} .$.
Similarly, $\angle MOE +\angle MEO +\angle OME =180^{\circ}$
$\Rightarrow \angle MOE+\angle MEO+90^{\circ}=180^{\circ}$
$\angle MOE+\angle MEO=90^{\circ} .$
From (i) and (ii) we get
$\angle LOE+\angle LEO=\angle MOE+\angle MEO$
Also, $\angle LEO =\angle MEO$ (Given) ...(iv)
From (iii) and (iv) we obtain
$\angle LOE=\angle MOE$
Now in triangles $OLE$ and $OME$
$\angle LEO=\angle MEO \text { (Given) }$
$\therefore \angle LOE=\angle MOE \text { (Proved above) }$
$EO=EO \text { (Common) }$
$\therefore$ by ASA congruence criterion we have:
$\triangle OLE \cong \triangle OME$
$\therefore OL=OM(\text { by } CPCT)$
Thus, chords $A B$ and $C D$ are equidistant from the centre $O$ of the circle. Since, chords of a circle which are equidistant from the centre are equal.
$\therefore AB=CD$
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