In the rectangular ABCD, AC and BD are two diagonal lines of the rectangle, and point P is a moving point on the side AD of the rectangular ABCD. The two side lengths AB and BC of the rectangle are 8 and 15 respectively. What is the sum of the distances from point P to the two diagonal lines AC and BD of the rectangle?

Solution: Let the intersection of two diagonal lines AC and BD be O and connect OP.

Point P is PM⊥AC of PN⊥BD, and M and N are vertical feet.

Because of the rectangular ABCD, AC=BD= 17.

OA=OB=OC=OD= 17/2

Therefore: s △ OAB = s △ OBC = s △ OCD = s △ ODA =1/4×15× 8 = 30.

Also: s △ ODA = s △ OPA+s △ OPD.

So: 1/2? OA？ PM+ 1/2？ OD？ PN=30

So: PM+PN = 120/ 17.

That is, the sum of the distances from point P to the two diagonal lines AC and BD of the rectangle is 120/ 17.