+26393 Question:
\(\text{Let $CA=b$} \\ \text{Let $CB=a$} \\ \text{Let $AB=c$} \\ \text{Let $CM=l$} \\ \text{Let $\angle AMC = \varphi$} \\ \text{Let $\angle CMB = 180^\circ-\varphi$} \)
\(cos-rule:\\ \begin{array}{|rcll|} \hline \text{In }\triangle \text{ AMC} \\ \hline b^2 &=& l^2 + \frac{c^2}{4} - 2l\frac{c}{2}\cos(\varphi) \qquad (1) \\ \hline \end{array}\\ cos-rule:\\ \begin{array}{|rcll|} \hline \text{In }\triangle \text{ CMB} \\ \hline a^2 &=& l^2 + \frac{c^2}{4} - 2l\frac{c}{2}\cos(180^\circ-\varphi) \\ a^2 &=& l^2 + \frac{c^2}{4} + 2l\frac{c}{2}\cos(\varphi) \qquad (2) \\ \hline \end{array} \)
\(\begin{array}{|lrcll|} \hline (1)+(2)& b^2 + a^2 &=& l^2 + \frac{c^2}{4} - 2l\frac{c}{2}\cos(\varphi) +l^2 + \frac{c^2}{4} + 2l\frac{c}{2}\cos(\varphi) \\ & b^2 + a^2 &=& l^2 + \frac{c^2}{4}+l^2 + \frac{c^2}{4} \\ & b^2 + a^2 &=& 2l^2 + 2\frac{c^2}{4} \\ & b^2 + a^2 &=& 2l^2 + \frac{c^2}{2} \\ & CA^2 + CB^2 &=& 2CM^2 + \frac{AB^2}{2} \\ \hline \end{array}\)
+118687 Thanks Heureka,
Note that my answer is the same as Heureka's
In triangle ABC, show
\(CA^2+CB^2=2CM^2+\frac{AB^2}{2}\)
Consider the triangle as I have drawn it and using Cosine rule.
\(a^2=d^2+x^2-2\;d\:x\;cos(180-Q)\\ a^2=d^2+x^2+2\;d\:x\;cos(Q)\qquad\qquad (1)\\~\\ b^2=d^2+x^2-2\;d\:x\;cos(Q)\qquad\qquad (2)\\ add\\ a^2+b^2=2d^2+2x^2\\ \text{Substituting original notation}\\ CB^2+CA^2=2(\frac{AB}{2})^2+2CM^2\\ CA^2+CB^2=2CM^2+2(\frac{AB^2}{4})\\ CA^2+CB^2=2CM^2+\frac{AB^2}{2}\\\)
