SOLUTION: ABCD is a trapezium such that AB is parallel to CD and angle DAB= angle DBC. Prove that BD = {{{sqrt (AB*CD)}}}

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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
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


Let x be the measure of angles DAB and DBC; let y be the measure of angle ADB.

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.

And in triangle ABD, the measure of angle ABD is also (180-x-y) degrees.

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.

Then corresponding parts of similar triangles gives us the desired result:

AB%2FBD+=+BD%2FCD
%28BD%29%5E2+=+AB%2ACD
BD+=+sqrt%28AB%2ACD%29