Question

The side AB of a parallelogram ABCD is produced to X and the bisector of angle CBX meets DA produced and DC produced to E and F respectively. Prove that DE=DF=BA+BC
Please answer fast.

Answer 1

Answer:

From the figure attached with this answer shows a parallelogram ABCD

AB = CD and BC = AD

the side AB of a is produced to X and the bisector of angle CBX meets DA produced and DC produced to E and F respectively. Prove that DE=DF=BA+BC

Let <CBF =<FBX = y

To prove AB + BC = DF

Then <CFB = y (since <CFB  = <FBX are alternate interior angles)

BC = CF (since <CBF =<CFB)

Therefore,

AB + BC = DC + CF = DF ------(1)

To prove AB + BC =DE

<CBF = <AEB ( corresponding angles)

AE = AB (since <AEB = <ABE)

AB + BC = AE + AD = DE  ------(2)

combining (1) and (2) we get

DE=DF=BA+BC

Answer 2

Answer:

From the figure attached with this answer shows a parallelogram ABCD

AB = CD and BC = AD

the side AB of a is produced to X and the bisector of angle CBX meets DA produced and DC produced to E and F respectively. Prove that DE=DF=BA+BC

Let <CBF =<FBX = y

To prove AB + BC = DF

Then <CFB = y (since <CFB  = <FBX are alternate interior angles)

BC = CF (since <CBF =<CFB)

Therefore,

AB + BC = DC + CF = DF ------(1)

To prove AB + BC =DE

<CBF = <AEB ( corresponding angles)

AE = AB (since <AEB = <ABE)

AB + BC = AE + AD = DE  ------(2)

combining (1) and (2) we get

DE=DF=BA+BC