A rectangular sheet of paper ABCD, with |AB| > |AD|, is folded so that two diagonally opposite corners A and C coincides - and the crease thus formed has length 5*|AD|/4.

Find |AB|/|AD|.

Let E be a point on side AB such that segment DE
is parallel to the crease (perpendicular to AC).

Clearly, triangles ABC and DAE are similar
since corresponding sides are perpendicular.

Therefore,

             |DE|     |AC|
            ------ = ------
             |AD|     |AB|

                     or       

                         |AD|*|AC|
             |DE| = -------------
                             |AB|

Combining this with the given:

             |DE| = (5/4)*|AD|

gives

             |AB| = (4/5)*|AC|

 ===>   |AD| = (3/5)*|AC|

 ===>   |AB|/|AD| = 4/3

QED

Posted by Bractals on 2016-06-17 14:11:59