Any point on a perpendicular bisector is equidistant from the endpoints of the line segment.
Suppose CM is the perpendicular bisector of AB, and that M is the midpoint of AB.
Two triangles can be created by connecting points A and C, and B and C.
These triangles both have a right angle and one of the legs measures half of AB. They also share one leg, CM.
According to the SAS Congruence Theorem, the triangles are congruent. Thus, their hypotenuses are also congruent.
Therefore, any point on a perpendicular bisector is equidistant from the endpoints of the segment. This can be summarized in a two-column proof.
Statement | Reason |
AM≅MBAB⊥CM | Given |
△AMC≅△BMC | SAS congruence theorem |
AC≅BC | △AMC≅△BMC |
AC=BC | Definition of congruent segments |
Determine the perimeter of △ABC.
A midsegment is a segment that connects the midpoints of two of the sides of a triangle.
Since a triangle has three sides, there are three midsegments.
The segment that connects the midpoints of two sides of a triangle — a midsegment — is parallel to the third side of the triangle and half the length.
If DE is a midsegment of △ABC, DE∥AB and DE=21AB.
If DE is a midsegment of △ABC, DE∥AB and DE=21AB.
Draw an arbitrary triangle, △ABC in the coordinate plane so that A lies on the origin, and AB is horizontal and lies on the x-axis.
Since A lies on the origin, its coordinates are (0,0). Point B is on the x-axis, meaning its y-coordinate is 0. The remaining coordinates are unknown, and can be named a, b, and c. Then, the points are A(0,0),B(a,0),andC(b,c). Draw midsegment DE from AC to BC. By defintion, D is the midpoint of AC and E the midpoint of BC.
If the slopes of DE and AB are equal, the segments are parallel. Since AB was drawn as a horizontal line, its slope equals 0. To show that the slope of DE also equals 0, it can be shown that the y-coordinates of its points are the same. To do this, A and C and the midpoint formula can be used. y-coordinate of D:yD=2y1+y2⇔yD=20+c⇔yD=2cy-coordinate of E:yE=2y1+y2⇔yE=20+c⇔yE=2c The y-coordinates of D and E are both 2c. It follows that the slope of DE equals 0. Thus, since AB and DE have equal slopes, they are parallel.
Since both AB and DE are horizontal, their lengths are given by the difference of the endpoints' x-values. The x-coordinates of A and B are 0 and a, respectively. This gives AB=a−0=a. To find the x-coordinates of D and E, the midpoint formula is used. x-coordinate of Dx-coordinate of E:xD=2x1+x2⇔xD=20+b⇔xD=2b:xE=2x1+x2⇔xE=2b+a⇔xE=2b+2a The length of DE is given by subtracting the x-coordinates: DE=2b+2a−2b=2a. Since DE=2a is half of AB=a, the midsegment DE is half the length of AB.
Therefore, a midsegment of a triangle is parallel to the third side of the triangle and half the length.If a segment parallel to one of the sides in a triangle is drawn between the other sides, the segment divides the other two sides proportionally.
For △ABC, since DE∥AB, DCAD=ECBE.
This can be proven using congruent angles.
Since DE and AB are parallel, the corresponding angles ∠CDE and ∠CAB are congruent. Similarly, ∠CED and ∠CBA are congruent. Also, triangles △ABC and △CDE share the angle ∠C.
A median of a triangle is a line segment between the midpoint of one side and the opposite vertex.
Since a triangle has three vertices, every triangle has three medians.
The three medians of a triangle intersect at one point called the centroid.
The centroid of a triangle is two-thirds of the distance from each vertex to the midpoint of the opposite side.
If AE, BD, and CF are the medians of △ABC, then AP=32AE,BP=32BD, and CP=32CF.
This can be proven using midpoints and parallel lines.
Consider △ABC. The points D and E are midpoints on their respective side. Thus, AE and BD are medians.
The two medians intersect at the point F.
Now, two new points are introduced — the midpoints of AF and BF. Call them G and H.
Since G and H are midpoints of AF and BF, GH is a midsegment of △AFB. Thus, by the Triangle Midsegment Theorem, GH is parallel to and half the length of AB.
Similarly, DE is a midsegment of △ABC since D and E are the midpoints of AC and BC. Therefore, DE is also parallel to and half the length of AB. It follows that GH∥DEandGH=DE.
Since DE and GH are parallel and congruent, DEHG are the vertices of a parallelogram. By the Parallelogram Diagonals Theorem, the diagonals of a parallelogram bisect each other. Therefore, DF≅FH and GF≅FE.
Thus, the median DB intersects AE at two-thirds of the distance from A. Now, by applying the same reasoning for the third median, it also intersects AE at two-thirds from A.
The median from point C also intersects AE at two-thirds from point A. Therefore, the centroid of a triangle is two-thirds the distance from each vertex to the midpoint of the opposite side.
Determine the length of AD.