333. The sum of the squares of two sides of a triangle is equal to twice the square of half the third side increased by twice the square of the median upon that side. Hyp. Let ABC be a A, AE the median bisecting BC. To prove AB+ AC2 = 2 BE2 + 2AE2. Proof. Draw AD other acute: let to BC. s AEB, AEC must be obtuse and the AEC be acute. 2 In the ▲ AEB, AB=BE2+AE + 2 BE× ED. (1) (331) In the ▲ AEC, AC2=CE2+AE’— 2 CE×ED. (2) Adding, and observing that BE = CE, we get 2 (330) AB2+ AC2= 2 BE2+ 2AE'. Q.E.D. 334. COR. Subtracting (2) from (1) in (333), we have AB2-AC2= 2 BC×ED. Hence, the difference of the squares of two sides of a triangle is equal to twice the product of the third side by the projection of the median upon that side. Let the student prove the case in which AD falls without the triangle ABC, or coincides with AE. NOTE.-By means of this theorem we may compute the medians of a triangle when the three sides are known. Proposition 29. Theorem. 335. If two chords cut each other in a circle, the product of the segments of the one is equal to the product of the segments of the other. 336. DEF. When four quantities, such as the sides about two angles, are so related that a side of the first is to a side of the second as the remaining side of the second is to the remaining side of the first, the sides are said to be reciprocally proportional. Therefore 337. COR. 1. If two chords cut each other in a circle, their segments are reciprocally proportional. 338. COR. 2. If through a fixed point within a circle any number of chords are drawn, the products of their segments are all equal. EXERCISES. 1. If AP 4, BP = 5, and CD = 12, find the lengths of CP and DP. = 2. If AB 20, and CD 24, find the lengths of CP and DP when CD bisects AB. Proposition 30. Theorem. 339. If from a point without a circle a tangent and a secant be drawn, the tangent is a mean proportional between the whole secant and the external segment. Hyp. Let PC and PB be a tangent and a secant drawn from the point P to the O ABC To prove PB: PC = PC: PA. Proof. Join CA and CB. In the AS PAC, PBC, Therefore, the square of the tangent is equal to the prod uct of the whole secant and the external segment. 341. COR. 2. Since PB × PA = - PC2, (340) PEX PD PC2, (340) Therefore, if from a point without a circle two secants be drawn, the product of one secant and its external segment is equal to the product of the other and its external segment. 342. COR. 3. If from a point without a circle any number of secants are drawn, the products of the whole secants and their external segments are all equal. Proposition 31. Theorem. fwo 343. If any angle of a triangle is bisected by a straight line which cuts the base, the product of the two sides is equal to the product of the segments of the base plus the square of the bisector. Hyp. In the ▲ ABC let AD bisect the Z BAC. To prove AB × AC=BD × DC+AD2. B Proof. Describe a O about the ABC, and produce AD to meet the Oce in E. Join CE. Then in the As BAD, EAC, D NOTE.-By means of this theorem we may compute the lengths of the bisectors of the angles of a triangle when the three sides are known. EXERCISE. If the vertical angle BAC be externally bisected by a straight line which meets the base in D, show that the product of AB, AC together with the square on AD is equal to the product of the segments of the base. Proposition 32. Theorem. 344. The product of two sides of a triangle is equal to the product of the diameter of the circumscribed circle by the perpendicular let fall upon the third side from the vertex of the opposite angle. Hyp. In the AABC let AD be to BC, and AE the diameter of the circumscribed O. Το prove AB × AC = AE × AD. Proof. In the AS ABD, AEC, and <B = <E, B each being measured by arc AC (238), E ZADB = /ACE = a rt. Z. (Cons.) and (240) NOTE.-By means of this theorem we may compute the radius of the circle circumscribed about a triangle when the three sides are known. EXERCISE. The product of the two diagonals of a quadrilateral inscribed in a circle is equal to the sum of the products of its opposite sides. |