Median to Hypotenuse of a Right Triangle

Problem : Let us consider the right triangle PQR with the right angle P (Figure 1), and let PS be the median drawn from the vertex P to the hypotenuse QR. We need to find the relationship between the length of the median PS and the the length of the hypotenuse QR.

Solution :Draw a straight line passing through the midpoint S and parallel to the side PR intersecting the side PQ at the point T. (Figure 2).

The angle QPR is given as right angle. The angles QTS and QPR are equal to each other as as they are corresponding angles of the parallel lines PR and TS and the transversal PQ. Hence the angle QTS is a right angle.

As TS passes through the mid-point S and is parallel to PR , it divides the side PQ into two equal parts i.e. PT = TQ. So, the triangles PTS and QTS are right triangle triangles with equal sides PT and TQ , these triangles also have a common side TS. Hence, these triangles are congruent in as per the Side – Angle – Side (SAS) Rule. 

From this we can say that the other sides of these triangles are also equal to each other as they are the corresponding parts of the congruent triangles , thus PS = QS. Now QS is equal to half the length of the hypotenuse QR , we can say that the median PS is also equal to half the length of the hypotenuse.

Hence, we can conclude that in a right triangle , the length of median to hypotenuse is half the length of the hypotenuse.

Locus of Mid Point of Falling Ladder

Locus: A locus of points is the set of points, and only those points, that satisfies given conditions. The locus of points at a given distance from a given point is a circle whose center is the given point and whose radius is the given distance.

Example : A 6-foot ladder is placed vertically against a wall, and then the foot of the ladder is moved outward until the ladder lies flat on the floor with one end touching the wall. What is the locus of the midpoint of the ladder as it slides?

Solution: The midpoint is on the hypotenuse of the right triangle whose legs are on the wall and floor. Since a right triangle can be inscribed in a semicircle with the midpoint of the hypotenuse (diameter of the circle) as its center, we know the distance OC is a radius of this circle and therefore 3 feet. That is, no matter where the ladder is, OC will be 3 feet, and therefore the locus of midpoints is a quarter of a circle with center at the intersection of the floor and wall (point O) and radius 3 feet.

In the following applet, press the play button at the lower left corner to see the various positions of the ladder as it is dragged and the movement of midpoint C of the ladder. You can also select the ‘Show Locus’ button to see the full locus and its equation. The geometrical solution to this problem can be accessed by selecting the ‘Solution’ button and then dragging the point A.

Right Triangle and Square on Hypetenuse

Let ABC be a right triangle with right angle at B. Let AC DE be a square drawn exterior to triangle ABC. If M is the center of this square, find the measure of ∠ MBC.
This is a Java Applet created using GeoGebra from www.geogebra.org - it looks like you don't have Java installed, please go to www.java.com

Note that triangle MCA is a right isosceles triangle with ∠AMC = 90° and ∠MAC = 45°. Since ∠ABC = 90°, there is a circle k with diameter AC which also passes through points B and C. Chord CM of circle k subtend angles MAC and MBC on the same segment . Hence ∠MBC = ∠MAC = 45°

Maximum Area of Right Triangle with given Hypotenuse

This is a Java Applet created using GeoGebra from www.geogebra.org - it looks like you don't have Java installed, please go to www.java.com