Triangle ABC has AB = 13, BC = 14, and AC = 15. Let P be the point on AC such that PC = 10. There are exactly two points, D and E, on line BP, such that quadrilaterals ABCD and ABCE are trapezoids. What is the distance DE?

The distance DE is `12 sqrt ( 2 ) .`

Expert Answers

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Refer to the attached image. For ABCD and ABCE to be trapezoids, they must have pairs of parallel sides. Draw the lines parallel to AB and BC until they intersect with PB to get the vertices D and E.

Then consider triangles APD and CPB. They are similar because all their corresponding angles are equal. And the coefficient of similarity is `AP / CP = 5 / 10 = 1 / 2 , ` so `AD = 1 / 2 BC = 7 . ` This implies that `DF = FA - DA = CB - DA = 7 , ` too.

Now we see that AD = DF, and the angles FDE and ADB, DFE and DAB are equal. This way, the triangles FDE and ADB are equal by ASA, so `FE = 13 .`

Further, the angles EFD and ABC are supplementary, so `cos ( EFD ) = - cos ( ABC ) . ` By the law of cosines,

`15^2 = 13^2 + 14^2 - 2 * 13 * 14 * cos ( ABC ) , ` `DE^2 = 13^2 + 7^2 + 2 * 13 * 7 * cos ( ABC ) ,`

so

`DE^2 = 13^2 + 7^2 + 1 / 2 ( 13^2 + 14^2 - 15^2 ) = 169 + 49 + 140 / 2 = 288 .`

`DE = sqrt ( 288 ) = 12 sqrt ( 2 ) .`

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