...understanding that they may be inserted as exercises for the student; some such possible exercises are: In any triangle, the product of any two sides is equal...segments of the third side formed by the bisector of the opposite angle, plus the square of the bisector. The medians of a triangle meet in one point which...

...three medians. Determine also the diameter of the circumscribed circle. PROPOSITION XXV. THEOREM 318. In any triangle, the product of any two sides is equal...segments of the third side formed by the bisector of the opposite angle, plus the square of the bisector. Hypothesis. AD bisects ZA of A ABC, meeting BC at...

...triangle the product of two sides is equal to the square of the bisector of their included angle, plus the product of the segments of the third side formed by the Abisector. Given : A ABC, CO bisector of Z ACS. To Prove : a • b = t* + n • r. Proof : Circumscribe...

...BCD. Ex. 51. From the conclusion of § 311, derive a formula for p* in terms of a, 6, and c. Ex. 52. In any triangle, the product of any two sides is equal...opposite vertex, minus the square of the bisector. D ,- ,. Prove AB x AC = DB x DC - AD2. " Suggestions. — 1. The solution is similar to that of §...

...drawn from C is 11. Find side BC. (§ 314. ) PLANE GEOMETRY — BOOK III PROPOSITION XXV. THEOREM 318. In any triangle, the product of any two sides is equal...segments of the third side formed by the bisector of the opposite angle, plus the square of the bisector. • tA. Give full proof. Hypothesis. AD bisects ZA...

...SUGGESTION. — To prove AC x BC= CD x CE. Draw BE. 42. In any triangle the product of two sides equals the product of the segments of the third side formed by the bisector of the opposite angle, plus the square of the bisector. SUGGESTION. — To prove AC x BC = ADx DB+ CD*. Circumscribe...