如图,在矩形ABCD中,AB=a,BC=2a,在BC上取一点P,使AB+BP=PD,求tan∠APD的值?

如图,在矩形ABCD中,AB=a,BC=2a,在BC上取一点P,使AB+BP=PD,求tan∠APD的值?
如图,在矩形ABCD中,AB=a,BC=2a,在BC上取一点P,使AB+BP=PD,求tan∠APD的值?

如图,在矩形ABCD中,AB=a,BC=2a,在BC上取一点P,使AB+BP=PD,求tan∠APD的值?
因为ABP和PCD都是直角三角形,故\x0dAB^2+BP^2=AP^2\x0dPC^2+CD^2=PD^2\x0d=> a^2 + x^2 = AP^2 ----- (1)\x0d(2a-x)^2 + a^2 = (a+x)^2 ---- (2)\x0d=> 由(2)得 x = 2/3*a ==> AP = square_root(13)/3*a\x0d又因为 角APD=180度-角APB-角DPC