310 



Solution of Mr. Harnett's Question. 



SOLUTION OF MR. HAMETt's OUESTION. BY MR. M. N. CRAW- 

 FORD. 



Cianford, Oct. 9, 1823. 



Let A B C be the given triangle right-angled at A. Upon 

 AB, AC, BC, describe the 

 squares AG, AD, and CH. 

 Then a right line being di'awn 

 from A parallel to BH will 

 meet the two lines G C, B D 

 in their common point of 

 intersection, which Mr, Ha- 

 rnett requires to be demon- 

 strated without the aid of 

 any proposition of the Ele- 

 ments beyond the 47th. 



Produce BC both ways to 

 meet perpendiculars on it from 

 G and D in M and N ; and 

 since the angle ABG is a right 

 angle, the sum of the angles 

 GBM, AB P is equal to a right 



angle (1 Elem. 13). Hence the angles at M and P being 

 right angles, and GB equal to BA, the triangles GBM, 

 BAP are identical (1 Elem. 26), and BM is equal to AP, 

 and GM to BP. In a similar manner it may be proved tlsat 

 the triangles DNC, CPA are identical, and that DN is equal 

 to CP, CN to AP and consequently to MB; and hence 

 CMtoBN. 



Complete the parallelogram BNDQ, and through d the 

 intersection of AT and BD draw TU parallel to BC, and 

 produce PA to II. Then because the complements Qd, rfN 

 are equal (1 Elem. 43), the parallelograms QP, UN are 

 equal ; that is, a parallelogram whose base and altitude are 

 Pf/, PN is equal to a parallelogram whose base and altitude are 

 B Q, B P or P C, B P. In the same manner it may be proved 

 that the parallelogram whose base and altitude are CM, P^ 

 (the segment of AP cut off by CG) is equal to the parallelo- 

 gram whose base and altitude are PC, BP. Consequently the 

 parallelogram whose base and altitude are Pd, B N is equal 

 to the parallelogram whose base and altitude are P^, C M. 

 Therefore, as CM, BN, have been shown to be equal, Fg is 

 equal to Fd; that is, the intersection of CG, AP, coincides 

 with the intersection of B D, A P. — Q. e. d. 



Mervyn Nott Crawford. 



