Question 1013605
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given: parallelogram ABCD, line DE is perpendicular to line AC, line BF is perpendicular to line AC
prove: triangle ADE is congruent to triangle CBF
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<pre>
0. Make a sketch. It is intended not to replace the proof, but to help you follow the proof.

1. Consider these two triangles, {{{DELTA}}}ADE and {{{DELTA}}}CFB.
   Notice that these triangles are right-angled triangles.

2. Their sides AD and BC are congruent as the opposite sides of the parallelogram.

3. These triangles have congruent angles <I>L</I>DAE and <I>L</I>FCB, since these angles are alternate interior angles 
   at the parallel lines AD and BA and the transverse AC.

4. Since the right angled triangles {{{DELTA}}}ADE and {{{DELTA}}}CFB have the pair of congruent acute angles <I>L</I>DAE and <I>L</I>FCB, 
   they have the other pair of acute angles <I>L</I>ADE and <I>L</I>CBF congruent too.

5. Now the triangles {{{DELTA}}}ADE and {{{DELTA}}}CFB are congruent according to the ASA-test of congruency of triangles.
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Regarding parallel line, see the lesson <A HREF=http://www.algebra.com/tutors/faq.mpl>Parallel lines</A> in this site.


Regarding tests on congruency of triangles, see the lesson <A HREF=http://www.algebra.com/algebra/homework/Triangles/Congruence-tests-for-triangles.lesson>Congruence tests for triangles</A>.