Question

in the given figure ABCD is a square and EF is parallel to the diagonal DB and EM=FM prove that BE=DE and AM bisect angle BAD

Answer 1

in ∆CBD,
as EF || DB
therefore CE/DE = CF/BF (BPT)
DE/CE = BF/CF
(DE+CE)/CE = (BF+CF)/CF
CD/CE = CB/CF
CE/CD = CF/CB
as CD = CB
therefore CE/CD = CF/CD
CE = CF
CD - CE = CB - CF
DE = BF
in ∆ABF and ∆ADE:
1) AB = AD
2) BF = DE
3) <ABF = <ADE = 90°
therefore ∆ABF congruent to ∆ADE
so AF = AE (CPCT)
also, <BAF = <DAE (CPCT) ___ (i)
in ∆AMF and ∆AME:
1) FM = EM
2) AM = AM
3) AF = AE
therefore ∆AMF congruent to ∆AME
so <MAF = <MAE (CPCT) ___ (ii)
from (i) and (ii):
<BAF + <MAF = <DAE + <MAE
<DAM = <BAM
therefore AM bisects <BAD