In trapezoid $ABCD$ with $AB$ parallel to $CD$ show that : $\frac{|AB|^2-|BC|^2+|AC|^2}{|CD|^2-|AD|^2+|AC|^2}=\frac{|AB|}{|CD|}=\frac{|AB|^2-|AD|^2+|BD|^2}{|CD|^2-|BC|^2+|BD|^2}$
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Tags: geometry
tk1
10.01.2017 01:40
The $LHS$ is a direct consequence of the law of cosines applied to triangles $BAC$ and $ACD$, since $\angle BAC=\angle ACD$. The $RHS$ is a direct consequence of the law of cosines applied to triangles $ABD$ and $BDC$, since $\angle ABD=\angle BDC$.
HadjBrahim-Abderrahim
10.01.2017 02:16
dangerousliri wrote: In trapezoid $ABCD$ with $AB$ parallel to $CD$ show that : $\frac{|AB|^2-|BC|^2+|AC|^2}{|CD|^2-|AD|^2+|AC|^2}=\frac{|AB|}{|CD|}=\frac{|AB|^2-|AD|^2+|BD|^2}{|CD|^2-|BC|^2+|BD|^2}$ tk1 wrote: The $LHS$ is a direct consequence of the law of cosines applied to triangles $BAC$ and $ACD$, since $\angle BAC=\angle ACD$. The $RHS$ is a direct consequence of the law of cosines applied to triangles $ABD$ and $BDC$, since $\angle ABD=\angle BDC$.
We must have $\measuredangle BAC=\measuredangle ACD\neq 90^\circ$ and, $\measuredangle ABD=\measuredangle BDC\neq 90^\circ.$ Otherwise, we have some problems because the first and the last fractions are not well-defined.
tk1
10.01.2017 02:44
Absolutely.