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Question:

Aug 10, 2021

#1
+26227
+3

Question:

$$\text{Let CA=b} \\ \text{Let CB=a} \\ \text{Let AB=c} \\ \text{Let CM=l} \\ \text{Let \angle AMC = \varphi} \\ \text{Let \angle CMB = 180^\circ-\varphi}$$

$$cos-rule:\\ \begin{array}{|rcll|} \hline \text{In }\triangle \text{ AMC} \\ \hline b^2 &=& l^2 + \frac{c^2}{4} - 2l\frac{c}{2}\cos(\varphi) \qquad (1) \\ \hline \end{array}\\ cos-rule:\\ \begin{array}{|rcll|} \hline \text{In }\triangle \text{ CMB} \\ \hline a^2 &=& l^2 + \frac{c^2}{4} - 2l\frac{c}{2}\cos(180^\circ-\varphi) \\ a^2 &=& l^2 + \frac{c^2}{4} + 2l\frac{c}{2}\cos(\varphi) \qquad (2) \\ \hline \end{array}$$

$$\begin{array}{|lrcll|} \hline (1)+(2)& b^2 + a^2 &=& l^2 + \frac{c^2}{4} - 2l\frac{c}{2}\cos(\varphi) +l^2 + \frac{c^2}{4} + 2l\frac{c}{2}\cos(\varphi) \\ & b^2 + a^2 &=& l^2 + \frac{c^2}{4}+l^2 + \frac{c^2}{4} \\ & b^2 + a^2 &=& 2l^2 + 2\frac{c^2}{4} \\ & b^2 + a^2 &=& 2l^2 + \frac{c^2}{2} \\ & CA^2 + CB^2 &=& 2CM^2 + \frac{AB^2}{2} \\ \hline \end{array}$$

Aug 11, 2021
#2
+80
+1

Hi! Thank you for answering my question. But could you tell me what the cosine rule is, please? Thank you!

jleung  Aug 11, 2021
#3
+26227
+3

Hello jleung,

heureka  Aug 12, 2021
#4
+115897
+1

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.

$$a^2=d^2+x^2-2\;d\:x\;cos(180-Q)\\ a^2=d^2+x^2+2\;d\:x\;cos(Q)\qquad\qquad (1)\\~\\ b^2=d^2+x^2-2\;d\:x\;cos(Q)\qquad\qquad (2)\\ add\\ a^2+b^2=2d^2+2x^2\\ \text{Substituting original notation}\\ CB^2+CA^2=2(\frac{AB}{2})^2+2CM^2\\ CA^2+CB^2=2CM^2+2(\frac{AB^2}{4})\\ CA^2+CB^2=2CM^2+\frac{AB^2}{2}\\$$

Aug 12, 2021