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
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Answers
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:
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