P is a moving point on the edge of rectangle ABCD, ab = 3, BC = 4, PM and PN are the distances to the diagonal respectively. Prove that (PM + PN) is a fixed value and calculate this value Did not learn the trigonometric function rectangle topic

P is a moving point on the edge of rectangle ABCD, ab = 3, BC = 4, PM and PN are the distances to the diagonal respectively. Prove that (PM + PN) is a fixed value and calculate this value Did not learn the trigonometric function rectangle topic

You didn't say which side it was on. I'll prove it on the AB side by pressing P. other things are the same
It is proved that PM and PN from P are vertical to AC and BD respectively
AC and BD intersect at point O and connect Po
AB = 3, BC = 4. Then AC = BD = 5
S rectangle ABCD = 12
The simple formula is s △ AOB = 1 / 4S rectangle ABCD = 3
S△APO=1/2×AO×PM,S△BPO=1/2×BO×PN
AO=BO=5/2
S△AOB=S△APO+S△BPO=1/2×5/2×（PM+PN）
Because the area of s △ AOB is constant, PM + PN is constant
The value is: 3 ÷ (1 / 2 × 5 / 2) = 12 / 5