a) Ta có:

{ˆABH=ˆBDC(AB//CD)ˆAHB=ˆBCD=900{ABH^=BDC^(AB//CD)AHB^=BCD^=900
 ⇒ΔABH∼ΔBDC(g.g)⇒ΔABH∼ΔBDC(g.g)

b) Ta có:

ΔABH∼ΔBDC(g.g)⇒AHBC=BHDC⇒AHBH=BCDC(1)ΔABH∼ΔBDC(g.g)⇒AHBC=BHDC⇒AHBH=BCDC(1)

Và:

CECE là tia phân giác của ˆBCDBCD^

⇒CBCD=EBED(2)⇒CBCD=EBED(2)

Từ (1),(2)⇒AHBH=EBED(1),(2)⇒AHBH=EBED

⇒AH.ED=BH.EB⇒AH.ED=BH.EB

c) Kẻ CK⊥BD=KCK⊥BD=K

Ta có:

SAECH=SAHE+SCHE=12HE.AH+12HE.CK(3)SAECH=SAHE+SCHE=12HE.AH+12HE.CK(3)

Mà:

⎧⎪ ⎪⎨⎪ ⎪⎩ˆAHB=ˆCKD=900AB=CDˆABH=ˆCDK(AB//CD)⇒ΔAHB=ΔCKD(ch−gn)⇒AH=CK(4){AHB^=CKD^=900AB=CDABH^=CDK^(AB//CD)⇒ΔAHB=ΔCKD(ch−gn)⇒AH=CK(4)

Từ (3),(4)⇒SAECH=AH.HE(∗)(3),(4)⇒SAECH=AH.HE(∗)

Lại có:

ΔABD;AB=8cm;AD=6cm⇒BD=√AB2+AD2=10cmΔABD;AB=8cm;AD=6cm⇒BD=AB2+AD2=10cm

Mặt khác:

ΔABH∼ΔBDC⇒AHBC=ABBD=BHDC⇒⎧⎪ ⎪ ⎪⎨⎪ ⎪ ⎪⎩AH=BC.ABBD=6.810=4,8cmBH=DC.ABBD=8210=6,4cm

Và:

EBED=CBCD=68=34⇒EB=37BD=307cm⇒EH=BH−EB=6,4−307=7435cm

Thay EH=7435cm;AH=4,8cm

SAECH=4,8.7435=1776175≈10,15cm2

Vậy SAECH≈10,15cm