M is the middle point of the side BC of the triangle ABC, and AB, AC are taken as the square ACDE and abgf to prove am = half EF

prove:
Extend am to point h, make MH = am, connect BH and ch
Then the quadrilateral abhc is a parallelogram
∴BH=AC=AE.∠ABH+∠BAC=180°
∵∠BAF=∠CAE=90°
∴∠EAF+∠BAC=180°
∴∠ABH=∠EAF
∵BH=AC=AE,AF=AB
∴△ABH≌△FAE
∴EF=AH=2AM
∴AM=1/2EF

It is known that △ ABC makes square abgf and ACDE outward with AB and AC as sides respectively. Point h is the midpoint of EF to prove ah ⊥ BC

Let AB = a (vector), AF = a ', AC = B, AE = B'
Then a & sup2; = a '& sup2;. AA' = 0, B & sup2; = B '& sup2;. BB' = 0, ab = - a'B '. A'B = ab' [all are number products]
AH＝（a'+b'）/2＝a'/2+b'/2,BC＝b-a.
AH·BC＝（a'/2+b'/2）·（-a+b）＝-a'a/2+a'b/2-b'a+b'b/＝0 ∴AH⊥BC

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