'Carpenter's square' seems an odd name for something L shaped 
but the key to its name, as to this 
tough nut, is in right angles. To prove that 

<math xmlns="http://www.w3.org/1998/Math/MathML">
<mrow><mo>&#x2220;</mo><mi>PBQ</mi><mo>=</mo><mo>&#x2220;</mo><mi>QBR</mi><mo>=</mo><mo>&#x2220;</mo><mi>RBC</mi></mrow></math> first 
draw the diagram with 
the carpenter's square placed so that R is on the 
line 
<math xmlns="http://www.w3.org/1998/Math/MathML">
<mrow><mi>DE</mi></mrow></math>, 
<math xmlns="http://www.w3.org/1998/Math/MathML">
<mrow><mi>P</mi></mrow></math> is on the line 
<math xmlns="http://www.w3.org/1998/Math/MathML">
<mrow><mi>AB</mi></mrow></math> and the edge 
<math xmlns="http://www.w3.org/1998/Math/MathML">
<mrow><mi>QT</mi></mrow></math> 
contains the point 
<math xmlns="http://www.w3.org/1998/Math/MathML">
<mrow><mi>B</mi></mrow></math>. Draw the lines 
<math xmlns="http://www.w3.org/1998/Math/MathML">
<mrow><mi>BQ</mi></mrow></math> and 
<math xmlns="http://www.w3.org/1998/Math/MathML">
<mrow><mi>BR</mi></mrow></math>. Then draw 
the line 
<math xmlns="http://www.w3.org/1998/Math/MathML">
<mrow><mi>RX</mi></mrow></math> perpendicular to 
<math xmlns="http://www.w3.org/1998/Math/MathML">
<mrow><mi>BC</mi></mrow></math> with 

<math xmlns="http://www.w3.org/1998/Math/MathML">
<mrow><mi>X</mi></mrow></math> on 
<math xmlns="http://www.w3.org/1998/Math/MathML">
<mrow><mi>BC</mi></mrow></math> and look for three congruent triangles.

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