1. If be, BC and EC are connected, then ∠ ACB = 90, ∴BC? =AB？ -AC? =28，cos∠BAC=AC/AB=3/4

∴cos∠bec=cos( 180-∠BAC)=-cos∠BAC =-3/4

Let EC = be = x (∵∠ BAE = ∠ EAC).

Is it derived from cosine theorem? +EC？ -2be×ECcos∠BEC = BC?

x？ +x？ +2x？ ×3/4=28

X=2√2 √ 2 is what you want.

2. link AC, BD

AB = cd arc ab = arc CD

∴ arc ab+ arc AC = arc CD+ arc AC

∴∠B=∠D

The quadrilateral ABDC is inscribed in a circle.

∴∠A+∠D= 180

∴∠A+∠B= 180

Minnesota ∴AC∥BD is the center line of trapezoidal ABDC.

∴MN∥AC

AMNC is an isosceles trapezoid.

∴∠AMN=∠CNM