document.write( "Question 696657: i dont understand how to do a 2 column proof or what to write inside the 2 column proof. can you help me? \n" ); document.write( "
Algebra.Com's Answer #429208 by MathLover1(20850)\"\" \"About 
You can put this solution on YOUR website!
A proof is just an orderly way to show that something is
\n" ); document.write( "true, by building on other things \"you+\"\"know\" are \"true\". \r
\n" ); document.write( "\n" ); document.write( "The only way that \"order\"\"+matters\" is that each thing you say \"must\" be based on something you've \"already+\"\"said\". Often it will be based on the previous statement,
\n" ); document.write( "but sometimes you will have to use earlier statements as well.\r
\n" ); document.write( "\n" ); document.write( " Your \"\"givens\"\" are the foundation someone laid for you, and the \"theorems\", \"properties\", and \"postulates\" you have to use to put together to a final goal, a \"proof\"\r
\n" ); document.write( "\n" ); document.write( "for example:\r
\n" ); document.write( "\n" ); document.write( ".Statements .......................... Reasons\r
\n" ); document.write( "\n" ); document.write( " 1. AB = CD ...................... given
\n" ); document.write( " 2. BC = BC ...................... reflexive property
\n" ); document.write( " 3. AB + BC = BC + CD ........ addition property of equality
\n" ); document.write( " 4. A-B-C-D ........................ given
\n" ); document.write( " 5. AC = AB + BC ................ definition of between
\n" ); document.write( " 6. AC = BC + CD ................ transitive property of equality
\n" ); document.write( " 7. BC + CD = BD ............. definition of between
\n" ); document.write( " 8. AC = BD ......... transitive property of equality\r
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\n" ); document.write( "\n" ); document.write( "so, here is the foundation at the bottom \"AC=BD\",and the goal at the top \"AB=CD\"\r
\n" ); document.write( "\n" ); document.write( "statement 2 gives you a new fact you can build on that doesn't depend on
\n" ); document.write( "the givens, but has been applied to a specific value, BC \r
\n" ); document.write( "\n" ); document.write( "statement 3 uses statements 1 and 2, by adding them together \"AB%2BBC=BC%2BCD\"\r
\n" ); document.write( "\n" ); document.write( "statement 4 is a given\r
\n" ); document.write( "\n" ); document.write( "statement 5 uses its definition \r
\n" ); document.write( "\n" ); document.write( "................\"3\"
\n" ); document.write( "...................|
\n" ); document.write( " \"AB%2BBC=BC%2BCD\"
\n" ); document.write( " ..... / ....... \
\n" ); document.write( " ....... 2 ....... 1\r
\n" ); document.write( "\n" ); document.write( "..... / .......... \
\n" ); document.write( " \"+BC=BC\" \"AB=CD\" \"+A-B-C-D\"\r
\n" ); document.write( "\n" ); document.write( "and you get \r
\n" ); document.write( "\n" ); document.write( "statement 7 is again based on the definition of between of statement 4
\n" ); document.write( "\"BC+%2B+CD+=+BD\"\r
\n" ); document.write( "\n" ); document.write( "finally we can join statements 6 and 7 to get to the goal which is \"AC+=+BD\"\r
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