Question

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Answer 1

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:

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

Answer 2

Answer:

this is your answer. look it once.