Question

梯形ABCD中,AD//BC,∠ABC=α（ ）,AB=DC=3,BC=5.点P为射线BC上动点（不与点B、C重合）,点

Answer 1

（1）∠PAB+∠ABC+∠APB=180°∠EPC+∠APE+∠APB=180°又∠ABC=α ∠APE=α所以∠PAB=∠EPC即∠1=∠2（2）当点P在线段BC上,即0＜X＜5时,∠1=∠2当X＞5时,∠2=∠APE+∠APB∠1=180°-（∠ABC+∠APB）又∠ABC=α ∠APE=...