Sta Znaci ANGLE BISECTOR na Hrvatskom - Hrvatski Prijevod

• simetrala kuta

Primjeri korištenja Angle bisector na Engleski i njihovi prijevodi na Hrvatskom

Let H be the point of intersection of the angle bisector and.
Let H biti točka na sjecištu simetrala kuta i.

Thus, the angle bisector of the angle CAB is the center line of these two circles.
Ovako, kut od simetrala kuta CAB je središnja linija tih dvaju krugova.

Length, is on the side, such that is the angle bisector of.
Duljina, je na strani, Kao da je simetrala kuta od.

Let the angle bisector of the angle OIH meet the Euler line e at a point W. Then,; since, we have.
Neka simetrala kuta u kut OIH susret Euler line e na jednom trenutku W. Tada,; Od, Imamo.

Prove that the line is the angle bisector of the angle..
Dokazati da je linija je simetrala kuta u kut.

Hence, the common tangent of these two circles at the point V is perpendicular to this angle bisector.
Dakle, zajednička tangenta tih dviju kružnica na točka V je okomita na ovu simetrala kuta.

Let be a point on the(internal) angle bisector of such that.
Dopustiti biti mjesto na(interne) simetrala kuta od tako da.

The center of this circle is simply the intersection of a normal to the line at the tangency point with the angle bisector.
Centar ovog kruga je jednostavno presjek normalan na liniju na tangens točka s simetrala kuta.

Hence, the point M lies on the angle bisector of the angle BAC.
Dakle, točka O leži na simetrala kuta u kut BAC.

The interior angle bisector of intersects the line in, and the perpendicular bisector of the side intersects the line in.
Unutrašnjost je simetrala kuta križa u retku u, A okomica simetrala od strane križa u retku u.

Let the parallel through to the interior angle bisector of intersect in.
Let kroz paralelne u unutrašnjosti od simetrala kuta presijecati u.

Since the angle bisector of an angle of a triangle divides the opposite side in the ratio of the adjacent sides, we have.
Budući da je simetrala kuta, kuta od trokut dijeli na suprotnoj strani u sistem u susjedstvu strane, imamo.

Similarly, the point N lies on the angle bisector of the angle BAD.
Slično tome, točka N leži na simetrala kuta u kut loše.

We will take the first two to be, where the conclusion is easy to draw, and we will take the third one such that is the angle bisector of.
Mi ćemo se prva dva da se, Gdje zakljucak je lako izvući, a mi ćemo se treći takav da je je simetrala kuta od.

Juniors 1 In a triangle with the incenter the angle bisector meets the circumcircle of triangle at point.
Juniors 1 U trokut s incenter je simetrala kuta zadovoljava circumcircle od trokut i točka.

Since the angle bisector of an angle of a triangle divides the opposite side in the ratio of the adjacent sides, we therefore have.
Budući da je simetrala kuta, kuta od trokut dijeli na suprotnoj strani u sistem u susjedstvu strane, mi, dakle, imaju.

From< PBC< DBA, you see that the line BD is the reflection of the line BP in the angle bisector of the angle ABC.
Od< PBC< DBA, vi vidite da u retku BD je odraz reda BP u simetrala kuta u kut ABC.

This is because the angle bisector of an angle in a triangle always passes through the midpoint of the arc cut off from the circumcircle by the opposite side.
To je zato što je simetrala kuta i kuta u trokut uvijek prolazi kroz midpoint od arc iskorijeni iz circumcircle prema suprotnoj strani.

Similarly, the point N' is the point of intersection of the angle bisector of the angle B'AD' with the line B'D', and we conclude that.
Slično tome, točka N'je točka na sjecištu simetrala kuta u kut B'AD' s line B would', a mi smo se zaključiti da.

In other words, the point R is the point of intersection of the perpendicular bisector of the segment MN with the angle bisector of the angle MAN.
Drugim riječima, na točku R je točka u sjecištu simetrala, okomica segment MN s simetrala kuta i kuta MAN.

But if an altitude and an angle bisector of a triangle coincide, then the triangle is isosceles; hence, the triangle GCH is isoscles, and the midpoint M of its base GH must lie on the altitude from the apex C. In other words, the point M lies on the line CI.
Ali ako visini i simetrala kuta nekog trokuta podudarati, tada je trokut jednakokračan, dakle, trokut GCH je isoscles, a midpoint M svojih baza GH mora počivati na nadmorskoj visini od apex C. Drugim riječima, točka M leži na liniji CI.

Let be the tangency point of the incircle with the triangle side and let be the intersections of the angle bisector with the incircle, so that the points follow on the angle bisector in this order.
Dopustiti biti tangens točka od incircle trokuta sa strane i neka biti na sjecištima je simetrala kuta s incircle, tako da točke slijedite na simetrala kuta u tom redu.

Therefore, if this common tangent intersects the line CA at a point V, then the triangle AVU is right-angled at V, and consequently,, since, as the point V, being the point of tangency of the circles k and, lies on their center line, i. e. on the angle bisector of the angle CAB.
Stoga, ako je ovo zajednička tangenta križa reda CA i točke V, onda je trokut AVU je pravo na V-angled, i time,, Od, Kao točka V, kao točka tangens od krugovima i k, Nalazi se na njihovom centru line, tj. na simetrala kuta u kut CAB.

Prove that is the bisector of angle. 3.
Dokazati da je od simetrala kuta. 3.

Let the bisector of the angle cut in.
Neka od simetrala kuta izrezati u.

To construct the center of the circle, we draw an arbitrary circle tangent to the lines centered on the bisector of the angle formed by the lines, which is identical with the bisector of the angle formed by the lines, i.e., with the internal bisector of the.
Da biste konstruirali centra u krug, Nacrtati krug proizvoljnog tangenta na linijama centered na simetrala kuta, formirana od strane linije, Koja je istovjetna s simetrala kuta, formirana od strane linije, Odnosno, s unutarnje simetrala od.

Is a triangle, the bisector of angle meets the circumcircle of triangle in, points and are defined similarly.
Je trokut, od simetrala kuta zadovoljava circumcircle od trokut u, Bodova i se definira na sličan način.

The bisector of angle meets the sides and of the quadrilateral at points and, respectively; the bisector of angle meets the sides and at points and, respectively.
The od simetrala kuta zadovoljava strane i od četverostrana na bodova i, Odnosno, od simetrala kuta zadovoljava strane i na boda i, Odnosno.

To construct the center of the circle, we draw an arbitrary circle tangent to the lines centered on the bisector of the angle formed by the lines, which is identical with the bisector of the angle formed by the lines, i.e.
Da biste konstruirali centra u krug, Nacrtati krug proizvoljnog tangenta na linijama centered na simetrala kuta, formirana od strane linije, Koja je istovjetna s simetrala kuta, formirana od strane linije, Odnosno.