Geometry: An Interactive Journey to Mastery



Solutions

5. /DEHOWKHIRXUYHUWLFHVRIWKHTXDGULODWHUDOA, B, C,
   and D as shown in Figure S.35.1, and regard these
   DVFRPSOH[QXPEHUV
   /HW¶VH[SUHVVHDFKRIP 1 , P 2 , P 3 , and P 4 in terms
   of A, B, C, and D.
   6WDUWLQJDWA, to reach P 1 ZHPXVW¿UVWZDONKDOIZD\
   along ABJJJGDQGWKHQWXUQFRXQWHUFORFNZLVHƒDQG
   walk the same distance again.
   Thus,
   P A ABiAB A 1     2211 JJJG JJJG 21 BA 2 iii BA^11  22 A B.
   In the same way,
   P 3   C^111 22 2 2CDJJJG JJJGiiiCD C D.
   Thus, PPJJJJG 13 LVJLYHQE\
   PPJJJJG 13     P 3 P 1 11112222 iiiiA B C D.
   In the same way,
   24 4 2

(^112222)
(^11112222).

PP P P

D AD iiAD B CB CB

iiiiABC D



§·§· ̈ ̧ ̈ ̧©¹©¹



JJJJG

Now, notice that

iPPiABC DJJJJG 24   ©¹ ̈ ̧§·^11112222  iiii^11112222 iiiiABC DPPJJJJG 13.

Thus, PPJJJJG JJJJG 13 isP P2 4URWDWHGFRXQWHUFORFNZLVHƒ&RQVHTXHQWO\WKHOLQHVHJPHQWVDUHFRQJUXHQW

and perpendicular.

P 1

A

B

D C

P 4

P 3

P 2

Figure S.35.1