In a parallelogram ABCD,AB=10cm and AD=6cm.The bisectors of <A meets DC in E,AF and BC produced meet at F.Find the length of CF
❌ Don't spam it ❌
Answers
Answer:
ANSWER
Given−
ABCDisaparallelogramwithAEForAFasthebisectorof
∠A.BCproducedmeetAFatF.AB=10cm,AD=6cm.
Tofindout−
ThelengthofCF=?
Solution−
AEisthebisectorof∠BAD
∴∠BAE=∠DAE.......(i)
ButAB∥DC(ABCDisaparallelogram)
∴∠BAE=∠DEA(altangles)
∠DAE=∠DEA(fromi)
∴ΔDAEisisosceles.⟹AD=DE=6cm⟹CE=DC−DE=(10−6)cm=4cm
AlsoAB=DC=10cm&AD=BC=6cm.(oppositesidesofaparallelogram).......(ii)
NowbetweenΔAFB&ΔEFCwehave
∠BAF=∠CEF&∠ABF=∠ECF(correspondingangles)
and∠AFBor∠EFCiscommon.
∴ΔAFB&ΔEFCaresimilar.
∴
CF
BF
=
EC
AB
⟹
CF
BC+CF
=
4
10
⟹
CF
BC
+1=
4
10
CF
6
+1=
4
10
⟹CF=4cm
Ans−CF=4cm
Step-by-step explanation:
mark me brainlist and follow me
❇️Answer ❇️
4cm
Step-by-step explanation:
Given−
ABCDisaparallelogramwithAEForAFasthebisectorof
∠A.BCproducedmeetAFatF.AB=10cm,AD=6cm.
Tofindout−
ThelengthofCF=?
Solution−
AEisthebisectorof∠BAD
∴∠BAE=∠DAE.......(i)
ButAB∥DC(ABCDisaparallelogram)
∴∠BAE=∠DEA(altangles)
∠DAE=∠DEA(fromi)
∴ΔDAEisisosceles.⟹AD=DE=6cm⟹CE=DC−DE=(10−6)
AlsoAB=DC=10cm&AD=BC=6cm.(oppositesidesofaparallelogram).......(ii)
NowbetweenΔAFB&ΔEFCwehave
∠BAF=∠CEF&∠ABF=∠ECF(correspondingangles)
and∠AFBor∠EFCiscommon
ΔAFB&ΔEFCaresimilar.
∴
CF
BF
=
EC
AB
⟹
CF
BC+CF
=
4
10
⟹
CF
BC
+1=
4
10
CF
6
+1=
4
10
⟹CF=4cm
Ans−CF=4cm
