document.write( "Question 969188: In parallelogram ABCD, AE is perpendicular to BC and AF is perpendicular to CD. Prove that triangle ABE is similar to triangle ADF. \n" ); document.write( "
Algebra.Com's Answer #592197 by Edwin McCravy(20056)\"\" \"About 
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document.write( "Given: parallelogram ABCD, AE ⊥ BC, AF ⊥ CD. \r\n" );
document.write( "To prove: ΔABE ∽ ΔADF.\r\n" );
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document.write( "1. m∠AEB = 90°      Given AE ⊥ BC.\r\n" );
document.write( "2. m∠AFD = 90°      Given AF ⊥ CD. \r\n" );
document.write( "3.  ∠AEB ≅ ∠AFD    Both have equal measures. 1,2\r\n" );
document.write( "4.    ∠B ≅ ∠D      Opposite interior ∠s of parallelogram ABCD are ≅.\r\n" );
document.write( "5.  ΔABE ∽ ΔADF    If 2 ∠s of one Δ are ≅ 2 ∠s of another Δ the Δs are ∽. 3,4\r\n" );
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document.write( "Edwin
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