Question

In a parallelogram ABCD the bisectors if A also bisects BC at x. If Ad =10cm, Find the value of AB.​

Answer 1

$\huge \sf \red{Given−}$

$\sf \blue{ABCD \: is \: a \: parallelogram \: with \: AEF \: or \: AF \: as \: the \: bisector \: of}$

∠A.BC produced meet AF at F. AB=10cm, AD=6cm}

$\sf \orange{To \: find \: out \: −}$

The length of CF=?

AE is the bisector of ∠BAD

∴∠BAE = ∠DAE..(i)

But AB ∥ DC (ABCD is a parallelogram)

∴∠BAE = ∠DEA(alt angles)

∠DAE=∠DEA(fromi)

∴ΔDAE isisosceles. ⟹AD = DE= 6cm ⟹CE =DC −DE = (10−6)cm =4cm

Also AB=DC = 10cm & AD=BC=6cm.(opposites idesofa parallelogram)...(ii)

Now between ΔAFB & ΔEFC we have

∠BAF = ∠CEF & ∠ABF = ∠ECF(corres ponding angles)

and ∠AFB or ∠EFC is common.

∴CFBF=ECAB ⟹ CFBC+CF=410

⟹CFBC+1=410

⟹CFBC+1=410CF6+1=410⟹CF=4cm

⟹CFBC+1=410 CF6 + 1 = 410

⟹CF = 4cm Ans − CF = 4cm

⟹CFBC+1=410CF6+1=410⟹CF=4cmAns−CF=4cm

Answer 2

Answer:

The length of CF=?

AE is the bisector of ∠BAD

∴∠BAE = ∠DAE..(i)

But AB ∥ DC (ABCD is a parallelogram)

∴∠BAE = ∠DEA(alt angles)

∠DAE=∠DEA(fromi)

∴ΔDAE isisosceles. ⟹AD = DE= 6cm ⟹CE =DC −DE = (10−6)cm =4cm

Also AB=DC = 10cm & AD=BC=6cm.(opposites idesofa parallelogram)...(ii)

Now between ΔAFB & ΔEFC we have

∠BAF = ∠CEF & ∠ABF = ∠ECF(corres ponding angles)

and ∠AFB or ∠EFC is common.

∴CFBF=ECAB ⟹ CFBC+CF=410

⟹CFBC+1=410

⟹CFBC+1=410CF6+1=410⟹CF=4cm

⟹CFBC+1=410 CF6 + 1 = 410

⟹CF = 4cm Ans − CF = 4cm

⟹CFBC+1=410CF6+1=410⟹CF=4cmAns