Elements of geometry and trigonometry . l be exi)lained more lully in another place. H I ) A / \ \ / 33 F 80 GEOMETRY. PROPOSITION XII. THEOREM. In every triangle, the square of a side opposite an acute angle isless than the sum of the squares of the other two sides, by twicethe rectangle contained by the base and the distance from theacute angle to the foot of the perpendicular let fall from theopposite angle on the base, or on the base produced. Let ABC be a triangle, and AD perpendicular to the baseCB ; then will AB-^AC^B0^—260 x CD. There are two cases. First. When the perpendicular falls
Elements of geometry and trigonometry . l be exi)lained more lully in another place. H I ) A / \ \ / 33 F 80 GEOMETRY. PROPOSITION XII. THEOREM. In every triangle, the square of a side opposite an acute angle isless than the sum of the squares of the other two sides, by twicethe rectangle contained by the base and the distance from theacute angle to the foot of the perpendicular let fall from theopposite angle on the base, or on the base produced. Let ABC be a triangle, and AD perpendicular to the baseCB ; then will AB-^AC^B0^—260 x CD. There are two cases. First. When the perpendicular falls withinthe triangle ABC, we have BDrrBC—CD,and consequently BD^-BCHCD^—2BCxCD (Prop. IX.). Adding AD- to each,and observing that the right angled trian-gles ABD, ADC, give ADHBD^=AB^, andADHCD^n^AC^ we have AB2-BC^+AC2—2BC X CD. Secondly. When the perpendicular ADfalls without the triangle ABC, we have BD=:.CD—BC ; and consequently BD-=CD^4-BC2—2CDxBC (Prop. IX.). ^ Adding AD^to both, we find, as before, AB=BC^+AC^— PROPOSITION XIII. THEOREM. In every obtuse angled triangle, the square of the side opposite t?ieobtuse angle is greater than the sum of the squares of the othertwo sides by twice the rectangle contained by the base and thedistance from, the obtuse angle to the foot of the perpendicularIçt fall from the opposite angle on the base produced. Let ACB be a triangle, C the obtuse angle, and AD perpen-dicular to BÇ produced ; then will AB2=ACHBC^-f2BCxCD. The perpendicular cannot fall within thetriangle ; for, if it fell at any point such asE, there would be in the triangle ACE, theright angle E, and the obtuse angle C, whichis impossible (Book L Prop. XXV. Cor. 3.) :
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