94 



JAMES CLERK MAXWELL. [CHAP. IV. 



x 4- 



Draw AC perpendicular to BC, and make an angle CAD = 

 HPX. D is the required point. If not, take any point E on the 

 opposite side of D from B, make AT = AE. Join TE, and make 

 DTL = right angle, then DTE > right angle DTL, because ATE 

 is less than a right angle (1.16), therefore L is within E. And in the 

 triangle DTL, DTL = PHX and TDL = HXP, therefore the triangles 

 are similar, and TD : DL = (XH : XP) = n:m; and n DL = m DT, then 

 %DE > mDT, add wBD and raTA or ?nEA, then riBT> + ?iDE + mEA > 

 ?iBD + mDT + mTA and nBD + raDA < wBE + mAE. 



Now take a point E- on the other side. Join AH, and cut off 

 AO = AD, make SRD = HXP = ADC. S will be beyond AR. Draw 

 DS perpendicular to AD, then it will be perpendicular to RS, and 

 below the line OD, then RSD similar to XHP and RS : RD = 

 (XH:XP = ) n:m, and m ~RS = n RD, but SR < RV < RO and 

 n RD ( = m RS) < m RO, add n BR and m OA or m DA, and 

 n BR + n RD + m DA < n BR + m RO + in OA and n BD + m DA 

 < n BR + m RA. QED. 



PROPOSITION 6 PROBLEM. 



To draw a tangent to an oval from a 

 focus without : 



Take m for the power of the greater 

 focus, and n for that of the less, and find 

 the angle ADB (Prop. 5). Upon AB de- 

 scribe a segment of a circle containing an 

 equal angle. Join B and C, the point where 

 it cuts the oval, B C is a tangent. For take 

 any point 0, join AO, m AO + n OB > 

 m AC + n CB, therefore is without the 

 oval. 



