Question

在矩形abcd中,ad=12,ab=7,de平分 adc,交bc于点e,ef⊥ae,交cd于点f

Answer 1

∠ABE = ∠ECF = 90 AB = EC =7

BE = 5 所以AE = 工74

∠BAE + ∠BEA = 90 ∠BEA + ∠CEF = 180 - ∠AEF = 90 所以∠BAE = ∠CEF

得△ABE与△CEF全等

AE = EF = 工74