For convenience, let F 1H 1=a,

A 1C 1=b，AD=c。

2

It is easy to know that ⊿A 1DC 1 is an isosceles right triangle.

∴(√2)a 1d=a 1c 1=2(？ A 1C 1)=2b。

That is, (√2)A 1D=2b.

∴A 1D=(√2)b.

three

According to the relationship between the topic and the inner and outer angles of the triangle:

∠a 1f 1D =∠ABD+∠c 1a 1f 1

=∠da 1c 1+∠c 1a 1f 1

=∠DA 1F 1。

That is ∠ a1f1d = ∠ da1f1.

∴ According to "equilateral", a1d = f1d.

∴F 1D=A 1D=(√2)b.

four

In the square ABCD, it is easy to know that BD = (√ 2) ad = (√ 2) c.

It is also easy to know BF1= (√ 2) f1h1= (√ 2) a.

And BD = BF1+f1d.

∴(√2)c=(√2)a+(√2)b.

∴c=a+b.

Namely: AD=F 1H 1+ (? )A 1C 1。