Question

Hey Friend Please Answer Fast
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Answer 1

Hey dude,



Here is your answer


Given:-- AB and CD are two chords of a circle,
With Centre O intersecting at a point E


Prove:-- AB=CD


Proof:-- angle LOE=180°-90°- angle LEO=90°- angle LEO [ angle some property ]

90°- angle AEQ=90°-angle DEQ
90°-angle MEO= angle MOE

In triangle OLE and OME,

Angle LEO=angle MEO

Angle LOE= angle MOE
(proved)

EO=EO (common)

OLE congruent OME

OL=OM ( cpct)

AB =CD (cpct)


Hope it is helpful for you ☯️☯️☯️☯️☯️



Thank you ☺️☺️☺️

Answer 2

$\mathbb{\red{\huge{Heya\:mate}}}$

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$\mathfrak{\blue{Solution.}}$Given that AB and CD are two chords of a circle, with center O intersecting at a point E.
PQ is a diameter passing through E, such that∠ AEQ = ∠ DEQ
Draw OL ⊥ AB and OM ⊥ CD.
In right angled ∆OLE
∠LOE + 90° + ∠ LEO = 180° (Angle sum property of a triangle)
∴∠LOE = 90° – ∠LEO 
= 90° – ∠AEQ = 90° – ∠DEQ
= 90° – ∠MEO = ∠MOE
In triangles OLE and OME,
∠LEO = ∠MEO 
∠LOE = ∠MOE (Proved)
OE = OE (Common side)
∴ ΔOLE ≅ ΔOME 
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$\mathbb{\huge{\green{Hope\:it\:helps!!}}}$