Dan je četverokut
![ABCD](/media/m/9/c/e/9ce25711ba18d9663b73c3580de4bf5a.png)
s kutovima
![\alpha = {60}^{\circ}](/media/m/8/0/1/801bd5d1676a954b7307a221773c252f.png)
,
![\beta = {90}^{\circ}](/media/m/6/1/e/61e7396601c713e8b4e2cf9dfe604d02.png)
,
![\gamma = {120}^{\circ}](/media/m/f/e/f/fef8e1af104669e7631cf149f0a95127.png)
. Dijagonale
![\overline{AC}](/media/m/d/9/5/d95354f0f833a5fda9c16a01a878c14f.png)
i
![\overline{BD}](/media/m/7/3/2/732e8894e57eb20026de06c47885ae55.png)
sijeku se u točki
![S](/media/m/c/6/3/c63593c3ec0773fa38c2659e08119a75.png)
, pri čemu je
![2\left\vert BS \right\vert = \left\vert SD \right\vert = 2d](/media/m/5/e/6/5e6122a5bc675405e4a7e67f31507673.png)
. Iz polovišta
![P](/media/m/9/6/8/968d210d037e7e95372de185e8fb8759.png)
dijagonale
![\overline{AC}](/media/m/d/9/5/d95354f0f833a5fda9c16a01a878c14f.png)
spuštena je okomica
![\overline{PM}](/media/m/4/6/4/464d7836c990c3d68f68a919b2f76146.png)
na dijagonalu
![\overline{BD}](/media/m/7/3/2/732e8894e57eb20026de06c47885ae55.png)
, a iz točke
![S](/media/m/c/6/3/c63593c3ec0773fa38c2659e08119a75.png)
okomica
![\overline{SN}](/media/m/f/5/9/f599515c303d29858a4968e44dfe6dd5.png)
na
![\overline{PB}](/media/m/7/4/7/747a83c57bdd49ea6ee905ebbcc0d534.png)
.
Dokaži:
a)
![\displaystyle \left\vert MS \right\vert = \left\vert NS \right\vert = \frac{d}{2}](/media/m/5/0/f/50f1d39f7d319e92d1022bb3fe09c239.png)
b)
![\left\vert AD \right\vert = \left\vert DC \right\vert](/media/m/8/3/f/83f41a5fc017d073f9d74edaa6235d5b.png)
c)
![\displaystyle P\!\left(ABCD\right) = \frac{9d^2}{2}](/media/m/a/d/d/add50a545ff38b234b93e753eead681f.png)
.
%V0
Dan je četverokut $ABCD$ s kutovima $\alpha = {60}^{\circ}$, $\beta = {90}^{\circ}$, $\gamma = {120}^{\circ}$. Dijagonale $\overline{AC}$ i $\overline{BD}$ sijeku se u točki $S$, pri čemu je $2\left\vert BS \right\vert = \left\vert SD \right\vert = 2d$. Iz polovišta $P$ dijagonale $\overline{AC}$ spuštena je okomica $\overline{PM}$ na dijagonalu $\overline{BD}$, a iz točke $S$ okomica $\overline{SN}$ na $\overline{PB}$.
Dokaži:
a) $\displaystyle \left\vert MS \right\vert = \left\vert NS \right\vert = \frac{d}{2}$
b) $\left\vert AD \right\vert = \left\vert DC \right\vert$
c) $\displaystyle P\!\left(ABCD\right) = \frac{9d^2}{2}$.