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The third question on page 59 of the ninth grade mathematics book published by Beijing Normal University.
∵ quadrilateral EFGH is a rectangle.

∴EF parallel to BC

∴∠AEF=∠ABC

∠AFE=∠ACB

∴△AEF similarity and△△△ ABC

∴AK/AD=EF/BC

∫AK = AD-KD = AD-EG

∴AD-EG

/

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=EF

/

B.C.

8-EG

/

eight

=EF/ 10

EF= 10*(8-EG)

/

eight

Let EG be X.

10*(8-x)

/

8 *

x

=

15

The next step is to solve the equation.

Hope to adopt. Thank you.