ABCD is a parallelogram. Pis the midpoint of side DC. Prove that 3AX = 2AC
Answers
Answer:
ABCD is a parallelogram and E is the midpoint of BC, such that if AB and it is extended they meet at point F.
\large {\boxed{\bigstar To \: Prove \: that \: DF = 2DC}}
★ToProvethatDF=2DC
In ΔBAE and ΔEFC
\boxed{ \mathsf{ \angle ABE = \angle ECF }}
∠ABE=∠ECF
\bold{REASON}REASON
Since, AD // BC (sides of parallelogram)
DAB and EBF are interior alternate angles.
\boxed{ \mathsf{ \angle AEB = \angle CEF}}
∠AEB=∠CEF
\bold{REASON}REASON
Vertically opposite angles.
\boxed{ \mathsf{ BE = EC}}
BE=EC
\bold{REASON}REASON
Since, E is the midpoint, it divides BC into equal parts.
\therefore \boxed{ \triangle ABE \cong \triangle ECF }∴
△ABE≅△ECF
by AAS property.
So, by C. P. C. T. C \boxed{AB = CF}
AB=CF
Now,
\bold{DF = DC + CF}DF=DC+CF
Since,
CF = AB (As proved above)
AB = DC (Opposite sides of parallelogram)
\therefore DC = CF∴DC=CF
So,
\mathsf{DF = DC + DC}DF=DC+DC
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