document.write( "Question 663949: Given: ABCD is a parallelogram, DE is perpendicular to AB, DF is perpendicular to BC.
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Algebra.Com's Answer #413070 by solver91311(24713) $\"\"$ $\"About$

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\n" ); document.write( "\n" ); document.write( "Angle DCF is congruent to Angle DAB because non-adjacent angles of a parallelogram are congruent. Angle DFC is right by defn perpendicular, as is Angle DEA. DF perpendicular to BC and BC parallel to AD means DF perpendicular to AD, hence Angle ADF is right. The measure of angle ADC is equal to the sum of the measures of angle ADF and angle FDC, hence the measure of angle FDC is equal to the measure of angle ADC minus 90. Similarly (you work out the details) the measure of angle ADE is equal to the measure of angle ADC minus 90. So the measure of angle ADE is equal to the measure of angle FDC. The triangles are similar by AAA. Q.E.D. \r
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\n" ); document.write( "\n" ); document.write( "John
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\n" ); document.write( "My calculator said it, I believe it, that settles it
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