In a trapizium ABCD AB||CD and DC=2AB EF || AB where E amd F lie on BC And AD respectively such that BE/EC=4/3.Diagonal DV intersects EF AT G.prove that 7 EF =11 AB.
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apply the trapezium formula
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given DC = 2 AB
Draw BH parallel to AD. H will be the middle point of DC as AB = DH.
Sin EF || AB, FI = AB
The ΔBHC and ΔBIE are similar as the corresponding sides are parallel.
So IE / HC = BE / BC
IE = HC * BE / BC
= y * 4x/7x = 4 y / 7
EF = FI + IE= y + 4 y /7 = 11 y / 7
so 7 EF = 11 AB
given DC = 2 AB
Draw BH parallel to AD. H will be the middle point of DC as AB = DH.
Sin EF || AB, FI = AB
The ΔBHC and ΔBIE are similar as the corresponding sides are parallel.
So IE / HC = BE / BC
IE = HC * BE / BC
= y * 4x/7x = 4 y / 7
EF = FI + IE= y + 4 y /7 = 11 y / 7
so 7 EF = 11 AB
Draw BH parallel to AD. H will be the middle point of DC as AB = DH.
Sin EF || AB, FI = AB
The ΔBHC and ΔBIE are similar as the corresponding sides are parallel.
So IE / HC = BE / BC
IE = HC * BE / BC
= y * 4x/7x = 4 y / 7
EF = FI + IE= y + 4 y /7 = 11 y / 7
so 7 EF = 11 AB
given DC = 2 AB
Draw BH parallel to AD. H will be the middle point of DC as AB = DH.
Sin EF || AB, FI = AB
The ΔBHC and ΔBIE are similar as the corresponding sides are parallel.
So IE / HC = BE / BC
IE = HC * BE / BC
= y * 4x/7x = 4 y / 7
EF = FI + IE= y + 4 y /7 = 11 y / 7
so 7 EF = 11 AB
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