document.write( "Question 1118690: ABCD is a trapezium such that AB is parallel to CD and angle DAB= angle DBC. Prove that BD = $\"sqrt+%28AB%2ACD%29\"$ \n" ); document.write( "

Algebra.Com's Answer #734090 by greenestamps(13203) $\"\"$ $\"About$

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\n" ); document.write( "Let x be the measure of angles DAB and DBC; let y be the measure of angle ADB.

\n" ); document.write( "Then since the sum of the measures of angles A and D in the trapezium is 180 degrees, the measure of angle BDC is (180-x-y) degrees.

\n" ); document.write( "And in triangle ABD, the measure of angle ABD is also (180-x-y) degrees.

\n" ); document.write( "So two angles of triangle ABD are congruent to two angles of triangle BDC, so the third angles are congruent, and triangles ABD and BDC are similar.

\n" ); document.write( "Then corresponding parts of similar triangles gives us the desired result:

\n" ); document.write( " $\"AB%2FBD+=+BD%2FCD\"$
\n" ); document.write( " $\"%28BD%29%5E2+=+AB%2ACD\"$
\n" ); document.write( " $\"BD+=+sqrt%28AB%2ACD%29\"$ \n" ); document.write( "

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