I need help, can someone do the whole proof, you will get good luck in return :}
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Question= Given M is a midpoint of BC ; M is the midpoint of AE. Prove that AB||CE
Answer = Given: ABCD is a parallelogram. M is mid point of BC.
In △DMC and △MBE
∠DMC=∠BME (Vertically opposite angles)
∠DCM=∠MBE (Alternate angles)
MC=MB (M is mid point of BC)
Thus, △DMC≅△EMB (ASA rule)
Thus, DC=BE (By CPCT)
Now, In △DPC and △APE
∠DPC=∠APE (Vertically Opposite angles)
∠CDP=∠AEP (Alternate angles)
∠PCD=∠PAE (Alternate angles)
Thus, △DPC∼△EPA (AAA rule)
Hence,
PD
PE
=
CD
AE
PD
PE
=
CD
AB+BE
(since AB = BE = CD)
PD
PE
=2
PE=2PD
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