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1971 AHSME Problems/Problem 5

Revision as of 00:10, 3 January 2019 by Guardian (talk | contribs) (Created page with "Solution 1: Let the measure of P equal (BD-AC)/2. BD = 80. Let the measure of Q equal 1/2 the measure of AC. Add the measures of P and Q to get (80-AC)/2 + AC/2 Therefore,...")
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Solution 1:

Let the measure of P equal (BD-AC)/2. BD = 80. Let the measure of Q equal 1/2 the measure of AC. Add the measures of P and Q to get (80-AC)/2 + AC/2 Therefore, the sum of P and Q is 40 as AC cancels out.

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