Theorem 422 Brocard Let ABCD be a quadrilateral inscribed in a circle with

# Theorem 422 brocard let abcd be a quadrilateral

This preview shows page 81 - 84 out of 238 pages.

Theorem 4.22 (Brocard) Let ABCD be a quadrilateral inscribed in a circle with center O and let E = AB CD , F = AD BC , and J = AC BD . Then O is the orthocenter of the triangle EFJ . Theorem 4.23 (The butterfly theorem) Let PQ be a chord of a circle and M be its midpoint . Let AB and CD be two other chords of the circle which pass through the point M . Let AD and BC intersect the chord PQ at the points X and Y , respectively . Then MX = MY . Theorem 4.24 (Maclaurin) Consider the angle xOy . Two points A and B are mov- ing on its sides Ox and Oy , respectively , in such a way that m OA + n OB = k, where m,n are given positive real numbers and k is a given straight line segment . Then the circumscribed circle of the triangle OAB passes through a fixed point . Remark 4.1 This point lies on the straight line Oh which is the geometrical locus of the points satisfying the property that their distances from the sides of the angle xOy are m/n . Theorem 4.25 (Pappou–Clairaut) Let ABC be a triangle . In the exterior of ABC , we consider the parallelograms ABB A , ACC A , P is the common point of the straight lines B A , C A , and T is the intersection point of AP , BC . On the extension of the straight line segment AT and in the exterior of the trian- gle , we consider a straight line segment TN = PA . Let BB C C be the parallelogram such that the point N belongs to the side B C . Then the equality S BB C C = S AA B B + S ACC A (4.26) holds true , where S denotes the enclosed area of the corresponding quadrilateral . Observation By using the Pappou–Clairaut Theorem, one can easily prove the Pythagorean Theorem. This can be done as follows (see Fig. 4.12 ). We observe that the right triangles HTA and ABC are equal. Indeed, it holds AC = AH and HT = AE = AB , 76 4 Theorems Fig. 4.12 Illustration of Pappou–Clairaut Theorem 4.25 hence AT = BC = BK , and therefore, BAM = A 1 = C. Thus A 2 + B = C + B = π 2 , where M is considered to be a point on the side BC which is the intersection point of the straight lines BC and AT . It follows that AT BC , which implies that AT BK . Since the squares are also parallelograms, by a direct application of the Pappou– Clairaut Theorem, we deduce that ( ABDE ) + ( ACZH ) = ( BKLC ), that is, AB 2 + AC 2 = BC 2 . 4 Theorems 77 Theorem 4.26 Let ABC be a triangle and M be a point in its interior . If x , y , z denote the distances of the point M from the sides BC , CA , and AB , respectively , of the triangle then the product xyz attains its maximum value when the point M is identified with the barycenter of ABC . Theorem 4.27 (Cesáro) Let the triangle ABC be given . Consider two triangles KLM and K L M circumscribed around the given triangle and similar to another given triangle with their respective homological sides perpendicular to each other , that is , KL K L , KM K M , ML M L . Then the sum of the areas of the triangles KLM and K L M , that is , S KLM + S K L M , is constant , where S denotes the enclosed area of the corresponding triangle .  #### You've reached the end of your free preview.

Want to read all 238 pages?

• • • 