McGraw Hill Integrated II, 2012
MH
McGraw Hill Integrated II, 2012 View details
6. Surface Areas and Volumes of Spheres
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Exercise 52 Page 855

E

Practice makes perfect

Let's analyze a segment AB and its bisector l at E.

Since l is a bisector, point E is in the middle of AB. This tells us that at least one point on l is in the same distance from point A as from point B. Now, let's choose a point X on line l, which is different than E.

Now, let's analyze △ AEX and △ EBX.

Since E is the midpoint of AB, we get that AE=EB. Furthermore, EX is a common side of △ AEX and △ EBX. Notice that the angles ∠ AEX and ∠ BEX are right. Therefore, by the Side-Angle-Side Similarity Theorem triangles △ AEX and △ EBX are congruent.

Since △ AEX and △ EBX are congruent, AX=BX. Therefore, point X is in the same distance from point A as from point B. This tells us that all points on l are the same distance from point A as from point B. This corresponds to option E.