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 20160617 14:11:59 