Mid-sections and angles

Question

In the triangle $ ABC $, whose $ AC> BC> AB $, on the sides $ BC $ and $ AC $ chose the point $ D $ and $ K $, respectively, so that $ CD = AB $, $ AK = BC$. Points $ F $ and $ L - $ midpoints $ BD $ and $ CK $ respectively. Points $ R,S - $ midpoints of $ AC $ and $ AB $ respectively. Segments $ CL $ and $ FR $ intersect at point $ O $, with $ \angle SOF = 55^{\circ} $. Find $ \angle BAC $.

Using GeoGebra, I received a reply ($ \angle BAC= 70^{\circ} $), but I can not prove it strictly

score 1 · Accepted Answer · answered Apr 07 '16 at 12:58

1

$$\varphi =55^\circ ,AB=DC,AK=BC,AS=SB,BF=FD,KL=LC$$ $$BC=KL=a, CA=b,AB=CD=c$$

$$BF=FD=\frac{a-c}{2},KL=LC=\frac{b-a}{2}$$

by Menelaus's theorem: $\frac{AN}{NB}\cdot \frac{a-c}{a+c}=1 \Rightarrow NB=\frac{a-c}{2}=BF$

by Menelaus's theorem: $\frac{BM}{MC}\cdot \frac{b-a}{b+a}=1 \Rightarrow CM=\frac{b-a}{2}=CL$

$SR - $ middle line $\Rightarrow$

$$ \varphi = \alpha+\beta,\angle A=180^\circ-2\varphi=70^\circ$$

answered Apr 07 '16 at 12:58

Roman83

17,884
3
26
70

Why do we have $B=2\alpha$ and $C=2\beta$ ? – Lucian Apr 09 '16 at 02:48
1

@Lucian: I can not understand you... $$\angle ABC = 180^{\circ}-\angle NBF$$ $$\angle NBF = 180^{\circ}-\angle BNF - \angle NFB$$ – Roman83 Apr 09 '16 at 07:57

Mid-sections and angles

1 Answers1