Bimedians and Diagonals in a Quadrilateral
Denote the midpoints of the sides \(AB\), \(BC\), \(CD\), and \(DA\) of quadrilateral \(ABCD\) as \(X\), \(Y\), \(Z\), \(W\), respectively. Segments \(XZ\) and \(YW\) are known as the bimedians of the quadrilateral. We have an engaging statement
The bimedians of a quadrilateral are equal if and only if its diagonals are perpendicular.
The bimedians of a quadrilateral are perpendicular if and only if its diagonals are equal.
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Copyright © 1996-2012 Alexander Bogomolny
The bimedians of a quadrilateral are equal if and only if its diagonals are perpendicular.
The bimedians of a quadrilateral are perpendicular if and only if its diagonals are equal.
Proof
The easiest way to establish the statement is by employing vector techniques. Choose an arbitrary origin \(O\) and associate with every point \(P\) the vector \(p=\overline{OP}\), denoted by the corresponding low case.
Thus we have, \(x=\frac{a+b}{2}\), \(y=\frac{b+c}{2}\), \(z=\frac{c+d}{2}\), \(w=\frac{a+d}{2}\). The statement at hand can be reformulated as
\(|x - z|=|y - w|\) iff \((a - c)\perp (b-d)\).
\((x - z)\perp (y - w)\) iff \(|a-c|=|b-d|\).
The absolute value of vector \(m\) is defined as the square root of the scalar product: \(|m|=\sqrt{m\cdot m}\); while the condition of orthogonality of two vectors \(m\) and \(n\) is also expressible in terms of the scalar product: \(m\perp n\) iff \(m\cdot n=0\). Using the scalar product we are able to once more reformulate our statement
\((x - z)(x - z)=(y - w)(y - w)\) iff \((a - c)(b-d)=0\).
\((x - z)(y - w)=0\) iff \((a-c)(a-c)=(b-d)(b-d)\).
Substituting \(x\), \(y\), \(z\), \(w\) from the definition reduces the above to
\((a+b - c-d)(a+b-c-d)=(b+c-d-a)(b+c-d-a)\) iff \((a - c)(b-d)=0\).
\((a+b - c-d)(b+c-d-a)=0\) iff \((a-c)(a-c)=(b-d)(b-d)\).
All that remains is pure algebra. For example,
\((a+b - c-d)(a+b-c-d)=(b+c-d-a)(b+c-d-a)\)
without the squares that are certainly the same on both sides, appears as
\(2(ab-ac-ad-bc-bd+cd)=2(bc-bd-ab-cd-ac+ad)\),
which is equivalent to \((a - c)(b-d)=0\).
Similarly, written explicitly, \((a+b - c-d)(b+c-d-a)=0\) gives
\(\begin{align}ab+ac & -ad-aa +bb+bc-bd-ab \\& -bc-cc+cd+ac -bd-cd+dd+ad \\& =2ac-aa+bb-2bd-cc+dd=0.\end{align}\)
which is equivalent to \((a-c)(a-c)=(b-d)(b-d)\).
Related material
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Copyright © 1996-2012 Alexander Bogomolny
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