Angles - Line in Space

Isosceles Trapeziums are Con-Cyclic

Problem : We need to prove that any isosceles trapezium is con-cyclic.

Solution :Here’s an isosceles trapezium with sides PQ and RS are parallel and PS = QR. A quadrilateral is cyclic if it’s opposite angles are supplementary (i.e. they add up to 180˚). So , we need to prove that QPS + QRS = 180˚ and PSR + PQR = 180˚.

Let us construct two perpendiculars, PU and QT, from point P and Q to segment RS.

Now, in ΔPSU and ΔQRT
PUS = QTR        by construction - both are right angles
PS = QR           Given trapezium is isosceles
PU = QT           perpendicular distance between two parallel lines
Thus, ΔPSU and ΔQRT are congruent by Right angle-Hypotenuse-Side (RHS) congruency rule
Now angles , PSU = QRT        Corresponding Parts of Congruent Triangles (CPCT)
Therefore ,angle PSR and angle QRS are equal. -------- (1)
Also, angle RQT is equal to angle SPU by CPCT. Adding right angles UPQ and TQP to the above angles, we get
RQT + TQP = SPU + UPQ
Thus, angle RQP is equal to SPQ -------- (2)
Adding equations 1 and 2 we get the following relations for angles
PSR + RQP = QRS + SPQ --------(3)
Since the sum of all the angles in a quadrilateral is 360˚,from equation (3)
PSR + RQP + QRS + SPQ = 360˚
2 (PSR + RQP) = 2 (QRS + QPS) = 360˚
PSR + RQP =QRS + QPS = 180˚
Since the opposite angles are supplementary, it can be concluded that an isosceles trapezium is a cyclic quadrilateral.

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 Problem - Feet of Perpendicular

Locus Problem : A ladder of length L is sliding over the floor. Find the locus of point D such that D is the foot of perpendicular dropped from point C such that OACB is a rectangle.

Solution: Let A (a,0) be the feet of the ladder and B (0,b) be the top of the ladder rest against the wall . O is the intersection point of floor and wall. OACB is a rectangle. D is the feet of the perpendicular drawn from point C onto ladder AB . As the feet of the ladder slides, the point D travels on a path called asteroid whose equation is given by x^(2/3)+y^(2/3)=L^(2/3).

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 point D of the ladder. You can also select the ‘Show Locus’ button to see the locus and its equation.

Locus of a General Point on Falling Ladder

In continuation to my previous post on locus of mid-point of a falling ladder , let us now explore the locus of a point lying anywhere on the falling ladder.

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 point which divides the ladder in the ratio 2 : 3 as it slides?

Solution: Let P be the point which divides the slider AB (A (a,0) being the feet and B (0,b) rest on the wall) .O is the intersection point of floor and wall.As the feet of the ladder slides,the point P whose coordinates are(3a/5,2b/5) travels on an elliptical path given by x²/9+y²/4=(6/5)².

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 point P of the ladder. You can also select the ‘Show Locus’ button to see the full locus and its equation.

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.

Nine Point Conics

Nine point circle of a triangle ABC is a circle passing through the mid points of sides of the triangle (G, H, I), feet of the perpendicular (D, E , F) drawn from the vertex to the opposite sides and the mid-points (K , L, M) of the distance from the orthocenter (N) to the three vertices of the triangle.

The concept of a nine point circle can be generalized to a nine point ellipse or a nine point hyperbola if we consider a general cevian instead of altitude. A cevian is any segment drawn from the vertex of a triangle to the opposite side. Cevians with special properties include altitudes, angle bisectors, and medians.

Consider three concurrent cevians with cevian point P, locate mid-points E, F and G of the segments from cevian point to the vertices of the triangle. Also locate the feet of the cevian K , N and L. If we draw a conic through any five of the above points, we will get an ellipse and it will also pass through the sixth point.

Now locate the mid points of the three sides of the triangle, we will find that these points also fall on the ellipse constructed above. The conic remains an ellipse when the feet of cevians lie on the sides of the triangle and converts to a nine point hyperbola when the feet of the cevians lie on the extensions of the sides.

Now locate centroid (G₁) of the triangle ABC and centre (N₁) of the conic , interestingly , the cevian point P , G₁ and N₁ lie on the same straight line with N₁P = 3 N₁G₁ . This is also the generalization of Euler Line.

Upper and Lower Sums