1973 AHSME Problems/Problem 35
In the unit circle shown in the figure, chords and are parallel to the unit radius of the circle with center at . Chords , , and are each units long and chord is units long.
Of the three equations
those which are necessarily true are
First, let be on circle so is a diameter. In order to prove that the three statements are true (or false), we first show that and then we examine each statement one by one.
Since by the Base Angle Theorem, By the Alternate Interior Angles Theorem, making by SAS Congruency. That means by CPCTC.
Because is a cyclic quadrilateral, , but we are given that is parallel to , so . Therefore, That makes an isosceles trapezoid, so
Lemma 2: Showing Statement I is (or isn’t) true
Let be the intersection of and By SSS Congruency, so We know that is a cyclic quadrilateral, so so That makes a parallelogram, so Thus,
In addition, and by the Base Anlge Theorem and Vertical Angle Theorem, That means by AAS Congruency, so .
By the Base Angle Theorem and the Alternating Interior Angle Theorem, so by ASA Congruency, Thus, Statement I is true.
Lemma 3: Showing Statement II is (or isn’t) true
From Lemma 2, we have . Draw point on such that making
Since we have Additionally, and so by SAS Congruency, That means
Since is an inscribed angle, Additionally, , so bisects Thus, making by SAS Similarity.
By using the similarity, we find that Thus, Statement II is true.
Lemma 4: Showing Statement III is (or isn’t) true
From Lemmas 2 and 3, we have and . Squaring the first equation results in Adding to both sides results in Since is positive, we find that which confirms that Statement III is true.
In summary, all three statements are true, so the answer is
It is a well-known fact that any cyclic trapezoid has its legs equal. Therefore, . Now, extend to meet the circle again at . By similar reasoning, . Furthermore, since and sum to and are equal, they have measure degrees. It trivially follows that . Dropping the altitude from to , we see that . Therefore, , , and .
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