Question

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
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Answer 1

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:

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Answer 2

❇️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