...is a conic. Reciprocate with respect to a circle the theorems contained in Exx. 5—12 inclusive. 5. The perpendiculars from the vertices of a triangle on the opposite sides are concurrent. 6. The tangent to a circle is perpendicular to the radius through the point of contact. 7. Angles in...

...+ CAB + ABC ; 2 (D AE + ED A) = ACB + D AC + D AE ; 2(DAE + DAC) = ACB + DAC + DAE ; .•.EAO=ECA. .•.BAC = ABC + BCA. § 20. The perpendicular bisectors...perpendicular to BC, AB. Draw AB', CB' perpendicular to AD, CF. Make AC'=AB'; CA' = CB'; and join C'A'; bisect C'A' in E. Then, since A, C, E, are the middle points...

...a conic. Reciprocate with respect to a circle the theorems contained in Exx. 5 — 12 inclusive. 5. The perpendiculars from the vertices of a triangle on the opposite sides are concurrent. 6. The tangent to a circle is perpendicular to the radius through the point of contact. 7. Angles in...

...Chapter X.] Reciprocate with respect to a circle the theorems contained in Exx. 5 — 12 inclusive. 5. The perpendiculars from the vertices of a triangle on the opposite sides are concurrent. 7. Angles in the same segment of a circle are equal. 8. The opposite angles of a quadrilateral inscribed...

...must show proficiency in both Part I. and Part II.) PART I. (GEOMETRY). 1. Prove that the respective perpendiculars from the vertices of a triangle on the opposite sides are concurrent. If P be the orthocentre of the triangle ABC, prove that the circum-circles of ABC, BPC, CPA, APB are...

...For conjugate diameters correspond to diameters at right angles in the circle (Prop. XXV., Cor.) and the perpendiculars from the vertices of a triangle on the opposite sides meet in a point. П. The locus of (M) the middle point of a focal chord in an ellipse is a similar...