Search results
Results from the WOW.Com Content Network
Consider a triangle ABC.Let the angle bisector of angle ∠ A intersect side BC at a point D between B and C.The angle bisector theorem states that the ratio of the length of the line segment BD to the length of segment CD is equal to the ratio of the length of side AB to the length of side AC:
The 'exterior' or 'external bisector' is the line that divides the supplementary angle (of 180° minus the original angle), formed by one side forming the original angle and the extension of the other side, into two equal angles. [1] To bisect an angle with straightedge and compass, one draws a circle whose center is the vertex. The circle ...
Draw the incenter by intersecting angle bisectors. Draw a line through I {\displaystyle I} perpendicular to the line A I {\displaystyle AI} , touching lines A B {\displaystyle AB} and A C {\displaystyle AC} at points D {\displaystyle D} and E {\displaystyle E} respectively.
An angle bisector of a triangle is a straight line through a vertex that cuts the corresponding angle in half. The three angle bisectors intersect in a single point, the incenter, which is the center of the triangle's incircle. The incircle is the circle that lies inside the triangle and touches all three sides. Its radius is called the inradius.
The center of the incircle, called the incenter, can be found as the intersection of the three internal angle bisectors. [3] [4] The center of an excircle is the intersection of the internal bisector of one angle (at vertex A, for example) and the external bisectors of the other two.
The intersection points of this circle with the two given lines (5) are T1 and T2. Two circles of the same radius, centered on T1 and T2, intersect at points P and Q. The line through P and Q (1) is an angle bisector. Rays have one angle bisector; lines have two, perpendicular to one another.
In a tangential quadrilateral, the four angle bisectors concur at the center of the incircle. [4] Other concurrencies of a tangential quadrilateral are given here. In a cyclic quadrilateral, four line segments, each perpendicular to one side and passing through the opposite side's midpoint, are concurrent.
Although originally formulated only for internal angle bisectors, it works for many (but not all) cases when, instead, two external angle bisectors are equal. The 30-30-120 isosceles triangle makes a boundary case for this variation of the theorem, as it has four equal angle bisectors (two internal, two external). [29]