In ∆ABC , AB = AC. D , E and F are the mid points of sides BC , CA and AB respectively.
Show that the line segment AD is perpendicular to the line segment EF and is bisected by it.
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Answer:
Given : AB = AC and D, E and F are mid points in BC, AB and AC respectively.
To Prove : AD is perpendicular to EF and is bisected by it
To construct : Join DE and DF.
Proof : Since EF are formed by joing mid points of triangle then EF || BC.
And since parallel lines are at 90° angle hence AD is perpendicular to EF.
Now, DF || AB and DF = 1 AB / 2
Since 1 AB / 2 = BE
DF = BE
Now, AB = AC
1 AB / 2 = 1 AC / 2
BE = AF
DF = AF
Now, taking triangle AOF and DOF.
DF = AF ( Proved above )
OF = OF ( Common )
angle AOF = ange DOF ( 90° each )
Hence triangle AOF ~ DOF by RHS congruency. Thus, AO = OD.
Hence EF bisect AD.
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