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.
✓
Answer
Given: AB and CD are two chords of a circle with centre O, intersecting at point E. PQ is a diameter through E, such that $\angle$AEQ = $\angle$DEQ.
To prove: AB = CD
Construction: Draw OL $\perp$ AB and OM $\perp$ CD
Proof: $\angle $LOE + $\angle $LEO + $\angle $OLE = 180° (Angle sum property of a triangle)
$\Rightarrow$ $\angle $LOE + $\angle $LEO + 90° = 180°
$\angle $LOE + $\angle $LEO =90° .........................(i)
Similarly, $\angle $MOE + $\angle $MEO + $\angle $OME = 180°
$\Rightarrow$$\angle $MOE + $\angle $MEO + 90° = 180°
$\angle $MOE + $\angle $MEO = 90° . .........................(ii)
From (i) and (ii) we get
$\angle $LOE + $\angle $LEO = $\angle $MOE + $\angle $MEO ..........(iii)
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 (Given)
$\therefore$ $\angle $LOE = $\angle $MOE (Proved above)
EO = EO (Common)
$\therefore$ by ASA congruence criterion we have:
$\triangle$OLE $\cong$ $\triangle$ OME
$\therefore$ OL = OM ( by CPCT)
Thus, chords AB and CD 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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