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ContentElementary Geometry
Elementary GeometryMiscellaneous Propositions953 If the inscribed circle of a triangle 𝐴𝐵𝐶 touches the sides 𝑎, 𝑏, 𝑐, in the points 𝐷, 𝐸, 𝐹; and if the escribed circle to the side 𝑎 touches 𝑎 and 𝑏, 𝑐 produced in 𝐷′, 𝐸′, 𝐹′; and if 𝑠=12(𝑎+𝑏+𝑐); then 𝐵𝐹′=𝐵𝐷′=𝐶𝐷′=𝑠−𝑐 and 𝐴𝐸′=𝐴𝐹′=𝑠 and similarly with respect to the other segments. ProofThe two tangents from any vertex to either circle being equal, it follows that 𝐶𝐷=𝑠−𝑐= half the perimeter of 𝐴𝐵𝐶, which is made up of three pairs of equal segments; therefore 𝐶𝐷=𝑠−𝑐 Also A𝐸′+A𝐹′=𝐴𝐶+𝐶𝐷′+𝐴𝐵+𝐵𝐷′ =2𝑠 therefore A𝐸′=A𝐹′=𝑠The NinePoint Circle954 The Ninepoint circle is the circle described through 𝐷, 𝐸, 𝐹, the feet of the erpendiculars on the sides of the triangle 𝐴𝐵𝐶. It also passes through the middle points of the sides of 𝐴𝐵𝐶 and the middle points of 𝑂𝐴, 𝑂𝐵, 𝑂𝐶; in all, through nine points.ProofLet the circle cut the sides of 𝐴𝐵𝐶 again in 𝐺, 𝐻, 𝐾; and 𝑂𝐴, 𝑂𝐵, 𝑂𝐶 in 𝐿, 𝑀, 𝑁. ∠𝐸𝑀𝐹=𝐸𝐷𝐹 (III. 21)=2𝑂𝐷𝐹 (952 Cor); therefore, since 𝑂𝐵 is the diameter of the circle circumscribing 𝑂𝐹𝐵𝐷 (III. 31), 𝑀 is the centre of that circle (III.20), and therefore bisects 𝑂𝐵.Similarly 𝑂𝐶 and 𝑂𝑂 are bisected at 𝑁 and 𝐿. Again, ∠𝑀𝐺𝐵=𝑀𝐸𝐷 (III.22)=𝑂𝐶𝐷, (III.21), by the circle circumscribing 𝑂𝐸𝐶𝐷. Therefore 𝑀𝐺 is parallel to 𝑂𝐶, and therefore bisects 𝐵𝐶. Similarly 𝐻 and 𝐾 bisect 𝐶𝐴 and 𝐴𝐵. 955 The centre of the ninepoint circle is the middle point of 𝑂𝑄, the line joining the orthocentre and the centre of the circumscribing circle of the triangle 𝐴𝐵𝐶. For the centre of the 𝑁.𝑃. cirlce is the intersection of the perpendicular bisectors of the chords 𝐷𝐺, 𝐸𝐻, 𝐹𝐾, and these perpendiculars bisect 𝑂𝑄 in the same point 𝑁, by (VI.2). 956 The centroid of the triangle 𝐴𝐵𝐶 also lies on the line 𝑂𝑄 and divides it in 𝑅 so that 𝑂𝑅=2𝑅𝑄. ProofThe triangles 𝑄𝐻𝐺, 𝑂𝐴𝐵 are similar, and 𝐴𝐵=2𝐻𝐺; therefore 𝐴𝑂=2𝐺𝑄; therefore 𝑂𝑅=2𝑅𝑄; and 𝐴𝑅=2𝑅𝐺; therefore 𝑅 is the centroid, and it divides 𝑂𝑄 as stated (951). 957 Hence the line joining the centres of the circumscribed and ninepoint circles is divided harmonically in the ratio of 2∶1 by the centroid and the orthocentre of the triangle.These two points are therefore centres of similitude of the circumscribed and ninepoint circles; and any line drawn through either of the points is divided by the circumferences in the ratio of 2∶1. See (1037) 958 The lines 𝐷𝐸, 𝐸𝐹, 𝐹𝐷 intersect the sides of 𝐴𝐵𝐶 in the radical axis of the two circles. For, if 𝐸𝐹 meets 𝐵𝐶 in 𝑃, then by the circle circumscribing 𝐵𝐶𝐸𝐹, 𝑃𝐸⋅𝑃𝐹=𝑃𝐶⋅𝑃𝐵; therefore (III.36) the tangents from 𝑃 to the circles are equal (985). 959 The ninepoint circle touches the inscribed and escribed circles of the triangl. ProofLet 𝑂 be the orthocentre, and 𝐼, 𝑄 the centres of the inscribed and circumscribed circles. Produce 𝐴𝐼 to bisect the arc 𝐵𝐶 in 𝑇. Bisect 𝐴𝑂 in 𝐿, and join 𝐺𝐿, cutting 𝐴𝑇 in 𝑆.The 𝑁.𝑃. circle passes through 𝐺, 𝐷, 𝐿 (954), and 𝐷 is a right angle. Therefore 𝐺𝐿 is a diameter, and is therefore = 𝑅=𝑄𝐴 (957). Therefore 𝐺𝐿 and 𝑄𝐴 are parallel. But 𝑄𝐴=𝑄𝑇, therefore 𝐺𝑆=𝐺𝑇=𝐶𝑇 𝐴2=2𝑅 𝐴2935,i Also 𝑆𝑇=2𝐺𝑆 𝑁 being the centre of the 𝑁.𝑃. circle, its radius=𝑁𝐺= 12𝑅; and 𝑟 being the radius of the inscribed circle, it is required to shew that 𝑁𝐼=𝑁𝐺−𝑟 Now 𝑁𝐼^{2}=𝑆𝑁^{2}+𝑆𝐼^{2}−2𝑆𝑁⋅𝑆𝐼 12𝑅−𝐺𝑆 𝑆𝐼=𝑇𝐼−𝑆𝑇=2𝑅 𝐴2−2𝐺𝑆 12𝐴, to prove the proposition. If 𝐽 be the centre of the escribd circle touching 𝐵𝐶, and 𝑟_{𝑎} its radius, it is shewn in a similar way that 𝑁𝐽=𝑁𝐺+𝑟_{𝑎}. Sources and Referenceshttps://archive.org/details/synopsisofelementaryresultsinpureandappliedmathematicspdfdrive©sideway ID: 210900021 Last Updated: 9/21/2021 Revision: 0 Ref: References
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