...sin 1? = e -r- sin (7, therefore pd = bc, qd — ca, rd = ab, and therefore generally — In every triangle the product of any two sides is equal to the product of the diameter of the circumscribing circle into the perpendicular on the third side from the opposite vertex....

...AD : AC, whence, ABXAC = AE X AD. PROPOSITION XXI.— THEOREM. 66. In any triangle, the product of two sides is equal to the product of the segments...the opposite angle plus the square of the bisector. Let AD bisect the angle A of the triangle ABC; then, ABX AC=DB X DC+ DA*. For, circumscribe a circle...

...AD : AC, whence, AB X AC = AE X AD. PROPOSITION XXI.—THEOREM. 66. In any triangle, the product of two sides is equal to the product of the segments...the opposite angle plus the square of the bisector. Let AD bisect the angle A of the triangle ABC; then, AB X AC= DBXDC + DA\ For, circumscribe a circle...

...AD. PROPOSITION XXI.— THEOREM. 66. In any triangle, the product of two sides is equal to the produd of the segments of the third side formed by the bisector...the opposite angle plus the square of the bisector. Let AD bisect the angle A of the triangle ABC; then, AB X AC=DBX DC+ DA2. For, circumscribe a circle...

...bisector. 8. The rectangle of any two sides of a triangle is equivalent to the rectangle of the external segments of the third side formed by the bisector of the opposite exterior angle, minus the square of the bisector. 9. The area of a triangle is equal to the product...

...CF to CE, and so also is PH to PB. PROPOSITION XVIII. THEOREM. 299. In any triangle the product of two sides is equal to the product of the segments...side formed by the bisector of the opposite angle together -with the square of the bisector. E Let ¿BAС of the ¿\ ABC be bisected by the straight...

...С F to CE, and so also is PH to PB. PROPOSITION XVIII. THEOREM. 299. In any triangle the product of two sides is equal to the product of the segments...third side formed by the bisector of the opposite anyle together with the square of the bisector. . ! Dí Let ¿BAC ofthe A ABC be bisected by the straight...

...F to СE, and so also is PH to PB. PROPOSITION XVIII. THEOREM. 299. 1n any triangle the product of two sides is equal to the product of the segments of the third side formed bi' the bisector of the opposite anyl e tor/ether with the square of the bisector. ~E Let Z. BA С...

...12, AD = 6, find diameter of the circumscribed circle. XL. Theorem. In any triangle, the product of two sides is equal to the product of the segments of the third side made by a line bisecting the opposite angle, plus the square of that line. HYPOTH. In A ABC, the line...

...the opposite side into segments proportional to the adjacent sides. In any triangle the product of two sides is equal to the product of the segments...the opposite angle plus the square of the bisector. Many other properties might be named as belonging to the triangle, such as those elicited by perpendiculars,...