Show that the position vector of E is

$\frac{1}{3}$ (a + b + c).

Explain briefly why this result shows that the three medians of a triangle are concurrent.

Show that the position vector f of F satisfies the equations

(b – f ) $\bullet$ (c – a) = 0

(c – f ) $\bullet$ (a – b) = 0.

Show, by expanding these equations, that

(a – f ) $\bullet$ (c – b) = 0.

Explain briefly why this result shows that the three altitudes of a triangle are concurrent.

d = b + $\frac{1}{2}$(c – b) = $\frac{1}{2}$(b + c)     (M1)A1

e = d + $\frac{1}{3}$(a – d)     M1

= $\frac{1}{2}$(b + c) + $\frac{1}{3}$(a - $\frac{1}{2}$(b + c))     A1

= $\frac{1}{3}$(a + b + c)     AG

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(because of the symmetry of the result), the other two medians also pass through E.     R1

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$\overrightarrow {{\text{BF}}}$ = f – b and $\overrightarrow {{\text{AC}}}$ = c – a     A1

since FB is perpendicular to AC, (b – f) $\bullet$ (c – a) = 0     R1AG

similarly since FC is perpendicular to BA, (c – f) $\bullet$ (a – b) = 0     R1AG

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expanding these equations and adding,     M1

b $\bullet$ c – b $\bullet$ a – f $\bullet$ c + f $\bullet$ a + c $\bullet$ a – c $\bullet$ b – f $\bullet$ a + f $\bullet$ b = 0     A1

– b $\bullet$ a – f $\bullet$ c + c $\bullet$ a + f $\bullet$ b = 0     A1

leading to (a – f) $\bullet$ (c – b) = 0     AG

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this result shows that AF is perpendicular to BC so that the three altitudes are concurrent (at F)     R1

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