document.write( "Question 387008: Prove lines containing the altitudes of any triangle are concurrent \n" ); document.write( "

Algebra.Com's Answer #273566 by richard1234(7193) $\"\"$ $\"About$

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Suppose triangle ABC has altitudes AD, BE, CF. Without loss of generality, let AD and BE meet at H. We want to prove that H lies on CF.\r
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\n" ); document.write( "\n" ); document.write( "On the contrary, suppose H is not on CF. Then, by Ceva's theorem,\r
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\n" ); document.write( "\n" ); document.write( " $\"%28AF%2FFB%29%28BD%2FDC%29%28CE%2FCA%29\"$ is not equal to 1 since the three segments are not concurrent.\r
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\n" ); document.write( "\n" ); document.write( "We can show that $\"CE%2FCD+=+BE%2FAD\"$ , $\"AF%2FAE+=+FC%2FBE\"$ , and $\"BD%2FBF+=+AD%2FFC\"$ using similarities within the right triangles ADC, AFC, etc. Multiplying all three equations,\r
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\n" ); document.write( "\n" ); document.write( " $\"%28CE%2FCD%29%28AF%2FAE%29%28BD%2FBF%29+=+%28BE%2FAD%29%28FC%2FBE%29%28AD%2FFC%29\"$ The right side of the equation cancels to 1, so\r
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\n" ); document.write( "\n" ); document.write( " $\"%28CE%2FCD%29%28AF%2FAE%29%28BD%2FBF%29+=+1\"$ \r
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\n" ); document.write( "\n" ); document.write( "However, the left side of the equation is equivalent to our first equation, so 1 is equal to some quantity other than 1, contradiction. Therefore the three altitudes must be concurrent. \n" ); document.write( "

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