An Interesting Right Triangle Property You’ll Need to Know for the GMAT

Quarter Wit, Quarter WisdomIn a previous post, we discussed medians, altitudes and angle bisectors of isosceles and equilateral triangles. Today, we will discuss an interesting property of perpendicular bisectors and circumcenter of right triangles.

Property: The circumcenter of a right triangle is the mid point of the hypotenuse.

Let’s prove this first and then we will see its application.

 

 

 

 

 

 

Say, we have a right triangle ABC right angled at B. Let’s draw the perpendicular bisector of AB which intersects AB at its mid point M. Say this line intersects the hypotenuse AC at N. We need to prove that AN = CN. Note that triangle AMN and triangle ABC are similar triangles using the AA property (angle AMN = angle ABC = 90 degrees and angle A is common to both triangles). So the ratio of the sides of the two triangle is the same. Since MN is the perpendicular bisector of line AB, AM = MB which means that AM is half of AB.

So AM/AB = 1/2 = AN/AC

Hence AN = NC

So N is the mid point of AC.

Using the exact same logic for side BC, we will see that its perpendicular bisector also bisects the hypotenuse. So N would be the circumcenter of triangle ABC and the mid point of AC.

Using an official question, let’s see how this property can be useful to us:

In the rectangular coordinate system shown above, points O, P, and Q represent the sites of three proposed housing developments. If a fire station can be built at any point in the coordinate system, at which point would it be equidistant from all three developments?

(A) (3,1)
(B) (1,3)
(C) (3,2)
(D) (2,2)
(E) (2,3)

 

 

 

 

 

First, let’s see how we will solve this question without knowing this property and using co-ordinate geometry instead.

Method 1:
Points O and Q lie on the X axis and are 4 units apart. We need a point equidistant from both O and Q. All such points will lie on the line lying in the middle of O and Q and perpendicular to the X axis. The equation of such a line will be x = 2. The fire station should be somewhere on this line.

Points O and P lie on the Y axis and are 6 units apart. We need a point equidistant from both O and P. All such points will lie on the line lying in the middle of O and P and perpendicular to the Y axis. The equation of such a line will be y = 3. The fire station should be somewhere on this line too.

Any two lines on the XY plane intersect at most at one point (if they are not overlapping). Since the fire station must lie on both these lines, it must be on their intersection i.e. at (2, 3).

This point (2,3) will be equidistant from O, Q and P. Therefore, the answer is E.

Method 2:
Think of the question in terms of the perpendicular bisectors of triangle OPQ. Their point of intersection will be equidistant from all three vertices.

We know that the circumcenter lies on the mid point of the hypotenuse. The end points of the hypotenuse are (4, 0) and (0, 6). The mid point will be

x = (4 + 0)/2 = 2
y = (0 + 6)/2 = 3

As in Method 1, the point (2, 3) will be equidistant from all three points, O, P and Q. Again, the answer is E.

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Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!