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20 10 the last question of mathematics in Jiaxing senior high school entrance examination
My supplement

20 10-07-03

10:56

Option d

(1) Because ACD and BCE are isosceles right triangles, BC/BE=AC/AD.

So BC*AD=BE*AC

Because AD and CE are parallel, CN/AD=BC/AB, which means BC*AD=AB*CN.

In the same way, BE*AC=CM*AB.

To sum up AB*CN=CM*AB.

So CN=CM

Because MCN angle =90 degrees.

So NMC angle = NCBI angle =45 degrees.

So MN is parallel to AB.

(2) BEcause CD is parallel to Be.

So triangle CDN is similar to triangle BEN.

So NC/EN=CD/EB.

So NC/CD=EN/EB

And because CD=AD, EC=EB.

So NC/AD=EN/EC

Because CN and AD are parallel.

MN is parallel to AB

So NC/AD=BC/AB, EN/EC=MN/AC.

So BC/AB=MN/AC.

That is BC/(AC+BC)=MN/AC.

So1/Mn = (AC+BC)/AC * BC =1/AC+1/BC.

My supplement

20 10-07-03

1 1:33

(3) Let the midpoint of AB be O, so AO=BO.

According to (2),1/Mn =1/AC+1/BC.

So MN=AC*BC/(AC+BC)=AC*BC/AB.

To prove that Mn

Only need to prove AC * BC/AB.

Because AC=AO+OC, BC=OB-OC=AO-OC.

So you only need to prove (ao+oc) * (ao-oc)/ab.

That is, ao 2-oc 2 < =1/4ab 2.

1/4ab^2=( 1/2ab)^2=ao^2

OC 2 > =0

Clearly established

So the original proposition proves that