Question

Two triangles ABC and AEF are such that B is the mid point of EF. If AB = EF = 1 and BC = 6, CA = 33 and
AB. AE + AC. AF = 2 then dot product of the vectors EF, BC is​

Answer 1

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.

\textbf{\large{ Q.E.D }}