Commentary on Proposition XIII.8¶

In order to prove this theorem we have to show (1) that the $\Delta\text{s}$ AEB, HAB are similar, and (2) that EH $= $ EA ($= $ AB).

To prove (2) we have

$\Delta\text{s}$ AEB, BAC equal in all respects,

whence

EB $= $ AC,

and

$\angle$ BAC $= \angle$ ABE.

Therefore

$\angle$ AHE $= 2\angle$ BAC $= \angle$ EAC,

so that

EH $= $ EA $= $ AB.

To prove (1) we have, in the $\Delta\text{s}$ AEB, HAB,

$\angle$ BAH $= \angle$ EBA $= \angle$ AEB,

and

$\angle$ ABE is common,

therefore the third $\angle\text{s}$ AHB, EAB are equal, and

$\Delta\text{s}$ AEB, HAB are similar.

Now, since these triangles are similar,

EB $:$ BA $= $ BA $:$ BH,

$(\text{rect. }$ EB, BH $) = (\text{sq. on }$ BA $)$ $= (\text{sq. on }$ EH $)$,

so that EB is divided in extreme and mean ratio at H.

Similarly its equal, CA, is divided in extreme and mean ratio at H.