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求：1.CE=2ED2.当PB=6cm时,求PD长

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
求：
1.CE=2ED
2.当PB=6cm时,求PD长

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