1963 AHSME Problems/Problem 18
Problem
Chord is the perpendicular bisector of chord
, intersecting it in
.
Between
and
point
is taken,
and
extended meets the circle in
. Then, for any selection of
, as described,
is similar to:
Solution
Note that is a right triangle with one of the angles being
. This leads to prediction that
is the similar triangle as it shares an angle, and to prove this, we need to show that
is a diameter.
If we let be on
so that
, then by SAS Congruency,
, so
. Since three points define a circle and point
is equidistant from three points,
is the center, so
is a diameter. Therefore,
is a right angle, and by AA Similarity, we can confirm that
. The answer is
.
See Also
1963 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 17 |
Followed by Problem 19 | |
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