Through D draw DHK parallel to AB. [I. 31.
Then each of the figures FH,HB is a parallelogram; therefore BH is equal to FG, and HK is equal to GB. [1.34.
Then, because HF is parallel to KC, [Construction.
one of the sides of the triangle DKC,
therefore KH is to HD as CE is to ED. [VI. 2.
But KH is equal to BG, and HD is equal to GF; therefore BG is to GF as GE is to ED. [V. 7.
Again, because FD is parallel to GE, [Construction.
one of the sides of the triangle AGE, therefore GF is to FA as ED is to DA. [VI. 2.
And it has been shewn that BG is to GF as CE is to EB. Therefore BG is to GF as CE is to ED, and GF is to FA as ED is to BA.
Wherefore the given straight line AB is divided similarly to the given divided straight line AC. q.e.f.
PROPOSITION 11. PROBLEM.
To find a third proportional to two given straight lines.
Let AB, AC be the two given straight lines: it is required to find a third proportional to AB, AC.
Let AB, AC be placed so as to contain any angle; produce AB, AC, to the points D, E; and make BD equal to AC; [I. 3.
join BC, and through D draw DE parallel to BC. [I. 31.
CE shall be a third proportional to AB, AC.
For, because BC is parallel to BE, [Construction.
one of the sides of the triangle ABE, therefore AB is to BD as AC is to CE; [VI. 2.
but BD is equal to AC; [Construction.
therefore AB is to AC as AC is to CE. [V. 7.
Wherefore to the two given straight lines AB, AC, a third proportional CE is found. q.e.f.
