Question

show that the perpendicular bisector of sides of a triangle using an activity with diagram
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Answer 1

Answer:

answer in explanation

Step-by-step explanation:

Let D,E,F be the midpoints of the sides BC,CA and AB of △ABC.

Let the perpendicular bisectors of the sides BC and AC meet each other in the points O. Choose O as the origin and let a¯,b¯,c¯,d¯,e¯,f¯ be the position vectors of the points A,B,C,D,E,F respectively.

Here we have to prove that OF¯¯¯¯¯=f¯ is perpendicular to AB¯¯¯¯¯=b¯−a¯.

By the midpoint formula, d¯=b¯+c¯2,e¯c¯+a¯2,f¯=a¯+b¯2

Now, OD¯¯¯¯¯=d¯ is perpendicular to BC¯¯¯¯¯=c¯−b¯.

∴d¯⋅(c¯−b¯)=0 ∴(b¯+c¯2)⋅(c¯−b¯)=0

∴(c¯+b¯)⋅(c¯−b¯)=0

∴c¯⋅c¯+b¯⋅c¯−c¯⋅b¯−b¯⋅b¯=0

∴c2−b2=0 ∴c2=b2...[∵c¯⋅c¯=c2,b¯⋅b¯=b2andc¯⋅b¯=b¯⋅c¯] . . .(1)

Also, OE¯¯¯¯¯=e¯ is perpendicular to AC¯¯¯¯¯=c¯−a¯

∴e¯⋅(c¯−a¯)=0

∴(c¯+a¯2)⋅(c¯−a¯)=0 ∴ as above c2=a2 . . . .(2)

From (1) and (2), we get,

b2=a2 ∴b2−a2=0

∴(b¯+a¯)⋅(b¯−a¯)=0

∴(b¯+a¯2)⋅(b¯−a¯)=0

∴f¯=OF¯¯¯¯¯ is perpendicular to b¯−a¯=AB¯¯¯¯¯.

∴ the perpendicular bisectors of the sides of ΔABC are concurrent.