Question

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: 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

*Random/irrelevant answers will not be tolerated !

Answer 1

See Below.

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:

Arc AB is the sum of Arcs AD and DB:

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

Since Arc AB ≅ Arc CD:

Solve:

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

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

Additionally:

Since they are vertical angles.

Thus:

By AAS.

Then by CPCTC:

Answer 2

this is your answer. look it once.