There are triangles ΔDEC and ΔBEF and a parallelogram ABCD in the given figure use the congruence property of triangles to prove CD = AB . We get from the parallelogram CD = AB. Then we use the given information. AF = 2AB. in the question to proceed.

Class: Grade 10

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Detailed Solution

Let ABCD is a parallelogram which means opposite sides are of equal length & opposite angles are equal.
AB = CD, AD = BC, ∠ADC = ∠ABC, ∠BAC = BCD.
E is midpoint of BC = BE = CE.
When DE & AB are extended they meet on a point F. RIP : AF = 2AB
from ΔDEC & ΔBEF ∠DEC & ∠BEF are vertically opposite to each other.
∠DEC = ∠BEF (VOA).
The line BC cuts the parallel line DC & AF & creates alternate angles ∠DCE & ∠FBE.
∠DCE = ∠FBE (alternate angle are equal)
From ΔDEC & ΔBEF That
∠CDE = ∠BFE, BE = CE, ∠DCE = ∠FBE
ΔDEC ΔBEF (by AAS congruence)
CD = BF but from the parallelogram
we have DE = AB so BF = AB
Now the point B on the line segment is AF
So, AF = AB + BF
putting the values BF = AB in about equation
AF = AB + BF AB AB = 2AB

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