Kamp '13 - Geometrija 7.
Dodao/la:
arhiva3. studenoga 2013. U trokutu
![ABC](/media/m/a/c/7/ac75dca5ddb22ad70f492e2e0a153f95.png)
je
![\angle ACB=90^{\circ} + \frac12 \angle CBA](/media/m/0/5/c/05c7804a281b42f0aecb9cbd2f3aab0e.png)
, a
![M](/media/m/f/7/f/f7f312cf6ba459a332de8db3b8f906c4.png)
je polovište
![|BC|](/media/m/9/7/d/97de44f615d4f287ccbca87f0f452862.png)
. Kružnica sa središtem u
![A](/media/m/5/a/e/5ae81275ee67d638485e903bdc0e9cde.png)
siječe
![BC](/media/m/5/0/0/5005d4d5eac1b420fbabb76c83fc63ad.png)
u
![M](/media/m/f/7/f/f7f312cf6ba459a332de8db3b8f906c4.png)
i
![D](/media/m/7/0/0/7006c4b57335ab717f8f20960577a9ef.png)
. Dokažite
![MD = AB](/media/m/1/f/b/1fb0868d497348e115a5653b35c819a8.png)
.
%V0
U trokutu $ABC$ je $\angle ACB=90^{\circ} + \frac12 \angle CBA$, a $M$ je polovište $|BC|$. Kružnica sa središtem u $A$ siječe $BC$ u $M$ i $D$. Dokažite $MD = AB$.
Izvor: Kamp 2013. - Geometrija, M. M.
Komentari:
ikicic, 26. srpnja 2016. 00:01
helen1c, 22. srpnja 2016. 14:08