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Mathematical proof of the first volume of grade three
The relationship is: MD=MF, MD⊥MF.

Proof 1: As shown in the figure, extend DM through CE to N and connect.

FD、FN .

∫ square ABCD, ∴AD∥BE, AD=DC

∴∠ 1=∠2。

AM = EM,∠ 3 = ∠ 4,

∴△ADM≌△ENM

∴AD=EN,MD=MN。

∵AD=DC,∴DC=NE。

∫ squared CGEF,

∴∠FCE=∠NEF=45,FC=FE,∠CFE=90 .

∫ squared ABCD, ∴∠ BCD = 90.

∴∠DCF=∠NEF=45,

∴△FDC≌△FNE。

∴FD=FN,∠5=∠6

≈cfe = 90 ,∴∠dfn=90 .

* DM = Mn,∴MD=MF,DM⊥MF.