Do ai and DH respectively, without vertical lines.

1, from the angle B45 degrees, AD//BC, AD=3, DC = 5ab = 4 2.

Get AI=BI=4/5, AD//BC, get DH=AI=4/5, HI=AD=3, then DH*DH+CH*CH=CD*CD? Get CH=28/5.

then what BC=BI+HI+CH=47/5

2. From the conditions BM=2CN and DH/CH=NO/CO= 1/7, NO=CO/7 is obtained.

CN*CN=CO*CO+NO*NO, so cn = 5 √ 2no-@ 1 (√ 2 stands for root number 2).

And MN//AB, the angle NMO = 45,

Available MO=CO/7? Then mo+co = BC-BM = BC-2cn-@ 2.

Are @ 1 and @2 at the same time? Get (8+ 10√2)NO=BC=47/5 Can you get NO? , CN = 5√2 No? That is, the value of time t (do it yourself)

3. There are two situations in which the triangle MNC is an isosceles triangle.

(1), when MN=CN, then we can get CO=MO from the second question.

CO+MO+BM=BC？ That is 2CO+2CN=BC, and CN = 5 √ 2no, and co = 7no to get CN.

(2) when CM=CN

CM+BM=BC, and BM=2CN? At the same time, CN=47/ 15=t

The figures may be miscalculated, and the thinking must be correct.

Dude, because I haven't done middle school math problems for n years, I'm still working so hard to write for you. Give points!