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☛[tex] \underline{ \underline{ \text{question}}}[/tex] : In the given figure , O is the centre of the circle. Two equal chords AB and CD intersect each other at E. Prove that :
i. AE = CE
ii. BE = DE

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  • xKelvin

    Answer:

    See Below.

    Step-by-step explanation:

    In the given figure, O is the center of the circle. Two equal chords AB and CD intersect each other at E.

    We want to prove that I) AE = CE and II) BE = DE

    First, we will construct two triangles by constructing segments AD and CB. This is shown in Figure 1.

    Recall that congruent chords have congruent arcs. Since chords AB ≅ CD, their respective arcs are also congruent:

    \stackrel{\frown}{AB }\, \cong\, \stackrel{\frown}{CD}

    Arc AB is the sum of Arcs AD and DB:

    \stackrel{\frown}{AB}=\stackrel{\frown}{AD}+\stackrel{\frown}{DB}

    Likewise, Arc CD is the sum of Arcs CB and DB. So:

    \stackrel{\frown}{CD}=\stackrel{\frown}{CB}+\stackrel{\frown}{DB}

    Since Arc AB ≅ Arc CD:

    \stackrel{\frown}{AD}+\stackrel{\frown}{DB}=\stackrel{\frown}{CB}+\stackrel{\frown}{DB}

    Solve:

    \stackrel{\frown}{AD}\, \cong\,\stackrel{\frown}{CB}

    The converse tells us that congruent arcs have congruent chords. Thus:

    AD\cong CB

    Note that both ∠ADC and ∠CBA intercept the same arc Arc AC. Therefore:

    \angle ADC\cong \angle CBA

    Additionally:

    \angle AED\cong \angle CEB

    Since they are vertical angles.

    Thus:

    \Delta AED\cong \Delta CEB

    By AAS.

    Then by CPCTC:

    AE\cong CE\text{ and } BE\cong DE