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20 16 Jiading district mathematics module 2
(1)∵ quadrilateral ABCD is a square,

∴∠ B =∠ ADC =∠ Bad = 90,AB = ad...( 1)

In △ABE and △ADF,

BE=DF∠B=∠ADF=90 AB=AD

∴ Abbe△ ADF (SAS) ... (1)

∴AE=AF,BAE = ∠ DAF...( 1)

∴∠eaf=∠ead+∠daf=∠ead+∠bae=∠bad=90。

AE = AF,

∴∠AFE=∠AEF.

∴∠AFE=∠AEF= 12×90 =45。

(2)∵ quadrilateral ABCD is a square,

∴∠ACB=∠ACD=45。

∫△ABE?△ADF,

∴∠AEB=∠AFD.

∠∠AEB =∠ACB+∠CAE = 45+∠CAE,∠AFD=∠AFE+∠CFM=45 +∠CFM,

∴∠CAE=∠CFM.

And material ≈ACB =∠ACD,

∴△ACE∽△FCM.

∴CECM=ACFC.