GEOMETRY. 



Since these triangles are similar, we have 



angle BCA equal to angle bca, 

 and angle BAG equal to angle bac, 



Again, join AD and ad. Then since the whole 

 angle BCD equals the whole angle bed, and the 

 angle BCA equals the angle bca, the remaining 

 angle ACD equals the remaining angle acd, and 

 the sides about the angles are proportional. 



BC . CA 



For -,- equals , 



be ^ fa' 



but 



BC , CD . 



-j-~ equals r, since the figures are similar; 



CA . CD 



.'. equals =-. 



ca cd 



Hence the triangles ACD, acd are similar. 

 The triangles AED, aed are also similar ; 

 .'. the figures are divided into the same number 

 of similar triangles. 



ABC AB 3 

 Again, 



and 



and 



abc 



ACD 

 acd 



AED 

 aed 



AC 3 = AB 3 

 ac* ab* ' 



AB 3 



ab* 



. ABC + ACD + AED = AB 2 

 abc + acd + aed ab 1 ' 



figure ABCDE = AB 3 BC 2 CD 3 

 figure abcde ~ ab* ~ be* ~ cd* ' 



PROP. XXXIV. In a right-angled triangle, 

 the perpendicular drawn from the right angle to 

 the hypotenuse divides the triangle into two tri- 

 angles, which are similar to the whole triangle 

 and to each other. 



Let ABC be a right-angled triangle, having the 

 angle at A a right 

 angle ; from A, AD 

 is drawn perpendic- 

 ular to BC ; it is 

 required to prove 

 that the two triangles 

 ABC, ACD are simi- B 

 lar to the whole tri- 

 angle and to each other. Comparing the two 

 triangles ABC and ABD, we have angle B com- 

 mon, angle BAC equal to angle ADB, each being 

 a right angle ; hence the triangles are similar. In 

 like manner, it may be shewn that the two tri- 

 angles ABC, ACD are similar ; therefore ABD, 

 ACD are similar to each other. 



Cor. I. Since ABC and ABD are similar, we 

 have 



BC AB 



AB ~ BD ; 

 /. AB is a mean proportional between BC and BD. 



Similarly, comparing the two triangles ABC, 

 ADC it may be shewn that AC is a mean pro- 

 portional between BC and CD ; and comparing 

 the two triangles ABD, ACD it may be shewn 

 that AD is a mean proportional between BD and 

 DC 



92 



Cor. 2. Since BC BD = AB 3 , 

 and BC CD = AC 2 ; 



/. BC{BD + CD} = AB 3 + AC 3 , 

 or BC 2 = AB 2 + AC 2 . 



Hence, in a right-angled triangle, the square on 

 the hypotenuse is equal to the sum of the squares 

 on the other two sides. 



PROP. XXXV. To find a mean proportional 

 to two given straight lines. 



Let A and B be the two given lines ; it is 

 required to find a mean proportional to them. 

 Take any straight line 

 CD ; from this line 

 cut off CE equal to A 

 and EF equal to B. 

 Upon CF describe a 

 semicircle. From E 

 draw EG perpendicu- 

 lar to CF, meeting the semicircle in G. Then EG 

 is a mean proportional to CE and EF. For join 

 CG, FG. Then the angle CGF is a right angle, 

 and GE is drawn from the right angle perpen- 

 dicular to CF ; then, by the last proposition, EG 

 is a mean proportional to CE and EF. 



PROP. XXXVI. To find a third proportional 

 to two given straight lines. 



Let A and B be the two given lines ; it is 

 required to find a third proportional to them. 

 Take any straight line CD ; 

 from this line cut off CE 

 equal to A, and EF equal 

 to B. At the point C draw a 

 line CG, making any angle 

 with CD ; and from CG 

 cut off CH equal to EF or 

 B ; and join EH, and draw 

 FK parallel to EH. Then HK is the line required 

 that is, HK is a third proportional to A and B. 



Since EH is parallel to FK, we have (Prop. 

 XXX.) 



CE CH 

 EF HK' 



or 



A B 

 B HK'' 

 .'. HK is a third proportional to A and B. 



PROP. XXXVI I. To find a fourth proportional 

 to three given straight lines. 



Let A, B, and C be the three given straight 

 lines ; it is required to 



find a fourth propor- M 



tional to them. Take 

 any line DE, and cut off 

 from it DF equal to A, 

 FG equal to B. At the 

 point D draw the line 

 DM, making any angle 



with DE ; cut off from DM, DH equal to C. 

 Join FH, and through G draw GK parallel to FH. 

 Then HK is the fourth proportional required. 



Since FH is parallel to GK, we have (Prop. 

 XXX.) 



DF _ DH 

 FG ~ HK' 



625 



