In a trapezium ABCD, AB parallel DCand DC= 2AB.EF parallel AB,where E and F lie on BC and ad respectively such that BE/EC=4/3. diagonal DB intersects EF at G prove that : 7EF=11AB.
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Answered by
393
see diagram.
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
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Answered by
75
Answer:7 EF =11 AB
Step-by-step explanation:
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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