Question

In the circle below, diameter CD is perpendicular to chord AB as shown. Radii OB and OA are drawn.

(a) How could we prove that OEB is congruent to OEA? Explain.




(b) What does your answer tell you must be true about segments AE and BE ? Why?


(c) If the diameter of circle O is 20 and CE  4, then determine the length of AB . Show how you arrived at your answer.

Answer 1

Answer:

The length of AB is 16 units.

Step-by-step explanation:

(a)

We know that $\angle OEB = 90\°$, by given, so it's a right angle, by definition.

$\angle OEA$ and $\angle OEB$ are adjacent angles on a straight angles. That means they are sumplementary angles by definition.

$\angle OEA + \angle OEB = 180\°\\ \angle OEA = 180\° -90\°\\\angle OEA = 90\°$

Therefore, $\angle OEA = \angle OEB$.

(b)

Sides AE and BE are congruent, because the congruence between those right angles, and the congruence between the other corresponding sides justify the congruence between sides AE and BE.

Notice that the hypothenuses of each right triangle are congruent, because they are also radius of the circle, which means they are congruent by definition.

Additionally, side OE is a common leg for both right triangles.

Therefore, sides AE and BE are congruent.

(c)

We know that $d=20$ and $CE=4$, that means the radius is

$r=\frac{d}{2}=\frac{20}{2}=10$

Notice that the chord AB is formed by AE and BE by sum of segments.

We know the hypothenuses are equal to 10 units of length, because they are radius.

Additionally,

$OC=OE+EC$, by sum of segments.

$10=OE+4\\OE=10-4\\OE=6$

Which means the common leg is 6 units long.

Now, we use the Pythagorean's theorem to find the missing leg

$OB^{2}=EB^{2}+OE^{2}\\ 10^{2}=EB^{2}+6^{2}\\100-36=EB^{2}\\EB=\sqrt{64}\\ EB=8$

But, $AB=AE+EB$, by sum of segments.

$AB=8+8=16$, because $AE=EB$ we said before.

Therefore, the length of AB is 16 units.