4(2) AADE = ACDF, ZADE = CDF
したがって,
ZEDF CDE + CDF
=ZCDE+ZADE
= ADC
4ARY=90°
線分DGは∠EDFの二等分線だから,
ZFDG = ½ZEDF x 90° = 45°
ZCDF
=x
FDG-Z CDG = 45° -20° = 25°
△CDFにおいて,
ZDFC 180° - (90° +25°) = 65°
(2)
=
AADE ACDF, AADE = ACDF
ABCD=AADE+BCDE
=ACDF+BCDE
=DEBF
=DEGF +ABEG.⑦
=
, ABEGƒÆABCD £ ŋ,
ABCD=6ABEG
⑦ より
DEGF =
....
de-
ABCD-ABEG
= 5ABEG ...
ここで, DEG と △DFGにおいて,
DE DF, DG = DG, ZEDG=/ FDG.
=
ADEGADFG
したがって, DFG
=
ウエより,
=
ADFG ABEG
ABEG ABEGO
=5:2
ADEG=ADFG
DEGF
A
E
H
ADFG=XFGX DC
ABEG = × EGX BH
よって,
1 G C
F
(x FGX DC) (EG × BH) = 5:2
FG = EG, DC=AB だから,
(EG × AB) (EG × BH) = 5:2
X
したがって, AB: BH = 5:2