Neocijenjeno
18. siječnja 2016. 01:12 (8 godine, 6 mjeseci)
Sakrij rješenje
U paralelogramu
![ABCD](/media/m/9/c/e/9ce25711ba18d9663b73c3580de4bf5a.png)
simetrala kuta
![\sphericalangle DAB](/media/m/f/d/8/fd81f94f045fb166610c69d8ae506717.png)
raspolavlja dužinu
![\overline{CD}](/media/m/3/3/8/338870e40f3ea7992d83158230115a5f.png)
. Ako sa
![M](/media/m/f/7/f/f7f312cf6ba459a332de8db3b8f906c4.png)
označimo polovište dužine
![\overline{CD}](/media/m/3/3/8/338870e40f3ea7992d83158230115a5f.png)
, odredite veličinu kuta
![\sphericalangle AMB](/media/m/2/6/2/262b2204f7c42814bb77c5bb097c91bc.png)
.
%V0
U paralelogramu $ABCD$ simetrala kuta $\sphericalangle DAB$ raspolavlja dužinu $\overline{CD}$. Ako sa $M$ označimo polovište dužine $\overline{CD}$, odredite veličinu kuta $\sphericalangle AMB$.
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Označimo
![\sphericalangle DAB=\alpha](/media/m/d/5/b/d5b5bf0e4071effa1d48f39ba5cdce39.png)
i
![\sphericalangle ABC=\beta](/media/m/7/2/b/72b84cfba1f9ed62473d56b999e4bf68.png)
. Budući da je
![ABCD](/media/m/9/c/e/9ce25711ba18d9663b73c3580de4bf5a.png)
paralelogram, znamo da vrijedi
![\alpha+\beta=180^{\circ}](/media/m/b/d/7/bd7d4287a628fa65763a57e47eb5e720.png)
. Prema pretpostavci zadatka je
![\sphericalangle DAM=\sphericalangle MAB=\frac{\alpha}{2}](/media/m/5/8/f/58f51d08dfbb359d5ad90c02d01d73e9.png)
i
![|CM|=|MD|](/media/m/3/9/8/3988cbb5170e78427cb43b2ded0e2182.png)
. No, vrijedi
![\sphericalangle AMD=\sphericalangle MAB=\frac{\alpha}{2}](/media/m/4/f/4/4f42e38c85e72bcb8eba8c307d1cac4d.png)
(jer su to kutovi uz presječnicu), pa je trokut
![MDA](/media/m/d/8/5/d85738c97a9361936fc041085f3520ee.png)
jednakokračan. Dakle,
![|MD|=|DA|=|BC|,](/media/m/3/6/a/36afa12420b16f62ccb11dd134dea679.png)
a odavde slijedi
![|BC|=|CM|](/media/m/7/3/5/7355579cb7a3ceb09f2489647c98ac98.png)
, tj. trokut
![BCM](/media/m/a/5/f/a5f96ee8a4e9b95ad0c0dd32593f7943.png)
također je jednakokračan. Zbog toga imamo
![\sphericalangle BMC=\sphericalangle MBC](/media/m/c/9/7/c97fa545b568c2f5b9f5701bad9e927a.png)
, a budući da također vrijedi
![\sphericalangle BMC=\sphericalangle MBA](/media/m/1/1/4/114d41fb29bf6bd866621840914fab32.png)
(jer su to kutovi uz presječnicu), slijedi
![\sphericalangle ABM=\sphericalangle CBM=\frac{\beta}{2}](/media/m/5/f/1/5f1ee7e48a3214802c84957304f1b691.png)
.\
Konačno imamo
![\sphericalangle AMB=180^{\circ}-\sphericalangle AMD-\sphericalangle BMC=180^{\circ}-\frac{\alpha+\beta}{2}=90^{\circ},](/media/m/f/2/1/f21e463ece2dce3d09eb6843668fb6df.png)
tj. kut
![\sphericalangle AMB](/media/m/2/6/2/262b2204f7c42814bb77c5bb097c91bc.png)
je pravi.
%V0
Označimo $\sphericalangle DAB=\alpha$ i $\sphericalangle ABC=\beta$. Budući da je $ABCD$ paralelogram, znamo da vrijedi $\alpha+\beta=180^{\circ}$. Prema pretpostavci zadatka je $\sphericalangle DAM=\sphericalangle MAB=\frac{\alpha}{2}$ i $|CM|=|MD|$. No, vrijedi $\sphericalangle AMD=\sphericalangle MAB=\frac{\alpha}{2}$ (jer su to kutovi uz presječnicu), pa je trokut $MDA$ jednakokračan. Dakle,
$$|MD|=|DA|=|BC|,$$
a odavde slijedi $|BC|=|CM|$, tj. trokut $BCM$ također je jednakokračan. Zbog toga imamo $\sphericalangle BMC=\sphericalangle MBC$, a budući da također vrijedi $\sphericalangle BMC=\sphericalangle MBA$ (jer su to kutovi uz presječnicu), slijedi $\sphericalangle ABM=\sphericalangle CBM=\frac{\beta}{2}$.\\
Konačno imamo
$$\sphericalangle AMB=180^{\circ}-\sphericalangle AMD-\sphericalangle BMC=180^{\circ}-\frac{\alpha+\beta}{2}=90^{\circ},$$
tj. kut $\sphericalangle AMB$ je pravi.