...the bars to be of the same thickness and density since the centroid of the perimeter of a triangle is the centre of the circle inscribed in the triangle formed by joining the middle points of the sides, if G be this centre, I) the middle point of AB, we have ¿ ADG = 90° - è (A -B). Now, for the equilibrium...

...to each other. 5. Shew that the centre of gravity of any open triangular truss, of equal scantlings, coincides with the centre of the circle, inscribed...triangle formed by joining the middle points of the side rafters and tie-rod of the given truss. 3" 6. Given an angle-iron of the dimensions, 6" x 4" x...

...to each other. 5. Shew that the centre of gravity of any open triangular truss, of equal scantlings, coincides with the centre of the circle, inscribed...triangle formed by joining the middle points of the side rafters and tie-rod of the given truss. 6. Given an angle-iron of the dimensions, 6" x 4" x --...

...contained by the sum and difference of the parts is equal to the rectangle contained by the parts. 3. The triangle formed by joining the middle points of the sides of a triangle is a fourth of the whole triangle. 4. The lines joining the middle points of the sides of...

...1 1. What is the analogous proposition in the First Book proving the equality of two triangles ? 2. The triangle formed by joining the middle points of the sides of another triangle is similar to that other. 3. Prove the proposition from the following construction...

...zero when the triangle becomes the Simson line. Let QRS, the pedal triangle, be denoted by A', and the triangle formed by joining the middle points of the sides of the triangle ABC by A ; the radius of the circle ABC, by R, and the distance between P and the centre of...

...to each other. 5. Shew that the centre of gravity of any open triangular truss, of equal scantlings, coincides with the centre of the circle, inscribed...triangle formed by joining the middle points of the side rafters and tie-rod of the given truss. 3" 6. Given an angle-iron of the dimensions, 6" x 4" x...

...mass is on the line DA, and at a distance from D equal to DA/3. (13) Show that the centre of mass of the triangle* formed by joining the middle points of the sides of a triangle has the same position as that of the latter triangle. (14) ABC is a triangle and D a fixed...

...mass is on the line DA, and at a distance from /> equal to DA]3. (13) Show that the centre of mass of the triangle* formed by joining the middle points of the sides of a triangle has the same position as that of the latter triangle. (14) ABC is a triangle and D a fixed...

...Prop. 6. — The centreso/T, T1? T3, T '3 coincide respectively with the incentre and the excentres of the triangle formed by joining the middle points of the sides of the orthocentric triangle o/ABC. Dem. — The line KN" is evidently the Simson's line of the point A' with...