Assignment 4

By: Olamide Alli

Exploration: Prove that the three perpendicular bisectors of the sides of a triangle are concurrent.

Step 1

We are given a triangle ΔABC. Construct the midpoint of AB, calling it M

Step 2
Construct the perpendicular bisectors, x, of segment AB.

Step 3
Construct a point on the perpendicular bisector x, calling the point D. Now construct two segments DA and DB.

Step 4

Because AN = NB, the angles DNA and DNB are equal. Since triangles ΔDNA = ΔDNB by the SAS theorem we can state the segments AD and DB are equal to each other, AD = DB.

Step 5

Construct the perpendicular bisector of BC, let’s call it y. WE know that the segments BC and AB are not parallel to each other, therefore we can say the perpendicular bisectors x and y intersect.

Step 6

Now combine the point D to the points of intersection for lines x and y. Therefore, the segments CG and GB are equal to each other (CG = GB). The angle CGD = angle BGD which implies ΔCGD = ΔBGD which means CD = DB. From the axioms from geometry we know that every segment is congruent to itself. Therefore DA = DC, DB = DA, CD = BC.

Step 7

Therefore D is on the perpendicular bisector of AC, which means that the perpendicular bisector of the sides of a triangle are congruent.

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