Question

如图在△ABC中,∠C=90°,D是BC边上一点,DE⊥AB于E ∠ADC=45°,DE:AE=1:5,若DE=根号2,

Answer 1

DE：AE=1：5

AE=5DE

DE⊥AB

AD^2 = AE^2+DE^2

AD=√(AE^2+DE^2) = √((5DE)^2+DE^2) = √26DE = √26*√2 = 2√13

∠ADC=45°

AC = ADsin45° = 2√13*√2/2 = √26