We will begin by drawing CE and EA, so that we will have two triangles â–ł EFA and â–ł EGC.
Our next step will be proving that △ EFA and △ EGC are congruent. Notice that both CE and EA are of ⊙ E. Therefore, since radii of a circle are congruent, CE and EA are also congruent.
Now, let's consider EF and EG. As we can see, both of them contain the center of the circle and they are to the chords. In this case, we should consider the (Theorem 10.7).
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Perpendicular Chord Bisector Theorem
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If a of a circle is perpendicular to a chord, then the diameter the chord and its .
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By this theorem, we say that EF and EG bisect AB and CD, respectively.
Since CD≅ AB, we can immediately conclude that CG≅ AF. Consequently, the and one leg of △ EFA are congruent to the hypotenuse and the corresponding leg of △ EGC. Therefore, by the , △ EFA and △ EGC are congruent.
△ EFA ≅ △ EGC
Since Corresponding Parts of Congruent Triangles are Congruent (CPCTC), EF and EG are also congruent. Thus, by the definition of congruent segments we can conclude that EF=EG.
Second Part of the Biconditional
For the second part, given that EF=EG we want to show that AB=CD.
We will again begin by drawing CE and EA, so that we will have two triangles â–ł EFA and â–ł EGC.
From the first part we know that CE and EA are congruent. Therefore, by the HL Theorem â–ł EFA and â–ł EGC are congruent. From here, by CPCTC we can conclude that CG and AF are congruent.
Now, as a result of Theorem 10.7, CG≅ DG and AF≅ BF.
Consequently, since AF≅ BF≅ CG ≅ DG, we can immediately conclude that AB≅ CD.
