Question

已知,正方形ABCD中,点E为AD边上一点,CE交对角线BD于点P,PE=AE

Answer 1

证明:(1)连AC,AP,AD=CD ∠ADP=∠CDP=45° DP=DP

⇒△ADP≅△CDP⇒PA=PC

⇒∠PAC=∠PCA

EA=PE⇒∠EAP=∠EPA=∠PAC+∠PCA

⇒∠EAP=2∠PAC

∠EAP+∠PAC=45°

⇒3∠PAC=45°⇒∠PAC=15°

⇒∠DAP=30°

△ADP≅△CDP⇒∠DCE=∠DAP=30°

⇒CE=2ED

(2)AD∥BC⇒ED/BC=PD/BD

tan∠DCE=ED/DC=1/√(3)

BC=DC=√(3 )ED

ED/√(3)ED=PD/6

∴PD=2√(3)cm