∵AC⊥BH, ∴AC slope is the negative reciprocal of BH, that is, -2.

∫A(5, 1), so the straight line of AC is y=-2x+ 1 1.

The equation of the straight line where the center line CM on AB side is located is 2x-y-5=0.

CM and AC are simultaneous, and the coordinates (4,3) of point C are obtained.

The equation of BH's straight line is x-2y-5=0.

You can set B(2w+5, w).

So the straight line where AB is located is: y = (w-1)/2w * (x-5)+1.

The method is: (x-5)/(2w+5-5) = (y-1)/(w-1).

The equation on which CM is online is 2x-y-5=0.

The abscissa of m is (7w+5)/(3w+ 1).

∵M is the midpoint of side AB, and the abscissa of∴ m is half of the abscissa of A and B, that is, (5+2w+5)/2.

Simultaneous solution w=0 (rounding) calculation process w is the denominator, or w=-3.

So B(- 1, -3)

Because c (4 4,3)

So BC: 6x-5y-9 = 0.