On the side BC of the triangle ABC we construct towards the exterior a square BCDE. Denote the intersection between AE and BC by M. Use the law of sines to prove that

$$\frac{BM}{CM}=\frac{\cos \measuredangle B\cdot \sin \measuredangle C}{\sqrt{2} \cdot \sin \measuredangle B \cdot \sin(\measuredangle C+45°)}$$

If someone could please help me prove this problem. I do not have a similar problem to work off of. I am unclear of where the $\sin(\measuredangle C+45^{\circ})$ comes into play and the $\sqrt{2}$.

  • $\begingroup$ the end got cut off...i am unlcear where the sin(<C+45) and \sqrt{2} come into play. $\endgroup$ – user423388 Mar 8 '17 at 13:42
  • $\begingroup$ Provide the figure $\endgroup$ – Nick Pavini Mar 8 '17 at 14:09

Let $\measuredangle B=\beta$, $\measuredangle C=\gamma$, $\measuredangle BAE=\alpha_1$, $\measuredangle CAE=\alpha_2$. By applying the sine rule to triangles $BAM$ and $CAM$ one gets: $$ {BM\over\sin\alpha_1}={AM\over\sin\beta},\quad {CM\over\sin\alpha_2}={AM\over\sin\gamma},\quad \hbox{whence:}\quad {BM\over CM}={\sin\alpha_1\over\sin\alpha_2}{\sin\gamma\over\sin\beta}. $$ By applying then the sine rule to triangles $BAE$ and $CAE$ one gets: $$ {\sin\alpha_1\over BE}={\sin(\beta+90°)\over AE},\quad {\sin\alpha_2\over CE}={\sin(\gamma+45°)\over AE}. $$ From that, taking into account that $CE=\sqrt2 BE$, one can readily obtain $\displaystyle{\sin\alpha_1\over\sin\alpha_2}$ and thus the desired result.

  • $\begingroup$ fully understood this !! thank you so much ! $\endgroup$ – user423388 Mar 8 '17 at 16:34

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Using sine rule at $ACE$ we have:

$$\frac{AE}{\sin(C+45°)}=\frac{AC}{\sin \alpha}$$

Using sine rule at $ABE$ we have:

$$\frac{AE}{\sin(B+90°)}=\frac{AB}{\sin \beta}$$

Dividing both equations we have:

$$\frac{\cos B}{\sin (C+45°)}=\frac{AC}{AB}\cdot \frac{\sin \beta}{\sin \alpha} \quad (1)$$

Using sine rule at $ABC$ we get

$$\frac{AC}{AB}=\frac{\sin B}{\sin C}$$

So, from $(1)$

$$\frac{\cos B\cdot \sin C}{\sin B\cdot \sin (C+45°)}=\frac{\sin \beta}{\sin \alpha}\quad (2)$$

but $\alpha + \beta=45°$ so

$$\frac{\sin \beta}{\sin \alpha}=\frac{\sin \beta}{\sin (45°-\beta)}=\sqrt{2}\cdot\frac{\tan \beta}{1-\tan \beta}\quad (3)$$

Finaly we can use, from the triangle BME, that

$$\tan \beta = \frac{BM}{BE}=\frac{BM}{BM+CM}\to \frac{BM}{CM}=\frac{\tan \beta}{1- \tan \beta} \quad (4)$$

Pluging $(3)$ and $(4)$ at $(2)$ we get what we want

  • $\begingroup$ can you explain (3)? i dont understand how you switched from sine to tangent... $\endgroup$ – user423388 Mar 8 '17 at 16:17
  • $\begingroup$ @erica: $\frac{\sin \beta}{\sin (45°-\beta)}=\frac{\sin \beta}{(\sqrt{2}/2)(\cos \beta-\sin \beta)}=$, now divide the numerator and denominator by $\cos \beta$ and get $\sqrt{2}\cdot\frac{\tan \beta}{1-\tan \beta}$ $\endgroup$ – Arnaldo Mar 8 '17 at 16:33

In triangle $ABM$, using the law of sines implies $$\frac{BM}{AM}=\frac{\sin\measuredangle BAM}{\sin\measuredangle B}$$ similarly, in $ACM$: $$\frac{CM}{AM}=\frac{\sin\measuredangle CAM}{\sin\measuredangle C}$$ combining these two yields: $$\frac{BM}{CM}=\frac{\sin\angle BAM}{\sin\angle CAM}\cdot\frac{\sin\angle C}{\sin\angle B}\label{*}\tag{*}$$ Now, in triangle $ABE$ note that $\measuredangle ABE=\measuredangle B+90^{\circ}$.

Thus $\sin\measuredangle ABE=\cos\measuredangle B$ and the law of sines implies $$\frac{\sin\measuredangle BAM}{\cos\measuredangle B}=\frac{BE}{AE}$$ and for triangle $ACE$: $$\frac{\sin\measuredangle CAM}{\sin(\measuredangle C+45)}=\frac{CE}{AE}$$ Note that $CE=\sqrt{2}BE$. Now combining the two equations implies: $$\frac{BE}{CE}=\frac 1{\sqrt 2}=\frac{\sin\measuredangle BAM}{\sin\measuredangle CAM}\cdot\frac{\sin(\measuredangle C+45)}{\cos\measuredangle B}$$ And finally, you can use $\eqref{*}$ to get the desired result.

  • $\begingroup$ OOps. I am late to the party $\endgroup$ – polfosol Mar 8 '17 at 15:00
  • $\begingroup$ can you explain how $\text{sin}\angleABE = \text{cos}\angleB$? $\endgroup$ – user423388 Mar 8 '17 at 15:58

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