In the figure, AD = BE, BC = DF and ∠ABC = ∠EDF. Prove that AC ïï EF and AC = EF.
Since AD = BE
... AD + DB = BE + DB
⇒ AB = DE
Now in ΔABC and ΔEDF
AB = DE ...... (proved above)
BC = DF ...... (given)
and ∠ABC = ∠EDF ...... (given)
... ΔABC ≅ ΔEDF
... AC = EF ...... (c.p.c.t.)
and ∠BAC = ∠DEF
but these are alternate interior angles of AC and EF with transversal AE
... AC ïï EF
... AD + DB = BE + DB
⇒ AB = DE
Now in ΔABC and ΔEDF
AB = DE ...... (proved above)
BC = DF ...... (given)
and ∠ABC = ∠EDF ...... (given)
... ΔABC ≅ ΔEDF
... AC = EF ...... (c.p.c.t.)
and ∠BAC = ∠DEF
but these are alternate interior angles of AC and EF with transversal AE
... AC ïï EF
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