Proof: As shown in Figure 2,1/AM+1/CK =1/en is known.

∫AB∨CD∨EF

∴∠

Let ∠ b = ∠ d = ∠ f = ∠ 1

Rt△ABM includes: sin ∠ b = sin ∠ 1 = am/ab.

Then: ab = am/sin ∠ 1

Similarly, in Rt△CDK and Rt△EFN, we can get:

CD = CK/sin∠ 1； EF=EN/sin∠ 1

∫1/am+1/CK =1/en (known)

(Both sides of the equation are multiplied by |1/symplectic1| at the same time.)

∴﹙ 1/sin∠ 1﹚﹙ 1/am+ 1/ck﹚=﹙ 1/sin∠ 1﹚﹙ 1/en﹚

Namely:1-am/Xin-1+1-CK/Xin-1=1.

∫ab = am/sin∠ 1； CD = CK/sin∠ 1； Ef = en/sin ∠ 1 (as proof)

∴ 1/a b+ 1/CD = 1/ef

(2) The relationship is:1/s △ Abd+1/s △ BDC =1/s △ bed.

Proof: ∫1/am+1/CK =1/en

∴﹙ 1/ bd﹚﹙ 1/am+ 1/ck﹚=﹙ 1/bd﹚﹙ 1/en﹚

(Both sides of the equation are multiplied by ﹙ 2/BD ﹚ at the same time, which can be obtained)

∴ 1/AM？ BD+ 1/CK？ BD= 1/EN？ Bachelor of science

∫s△Abd = am again? BD/2 → AM？ BD=2S△ABD

S△BDC=CK？ BD/2 → CK？ BD=2S△BDC

S△BED=EN？ BD/2 → EN？ BD=2S△BED

Substituting into the equation, we can get:

1/﹙2s△abd﹚+ 1/﹙2s△bdc﹚= 1/﹙2s△bed﹚

∴﹙ 1/2﹚﹙ 1/s△abd+ 1/s△bdc﹚=﹙ 1/2﹚﹙ 1/s△bed﹚

After omitting the constant 1/2 on both sides of the equation, we can get:

1/SδABD+ 1/SδBDC = 1/SδBED