The first part is finding $ \frac{AE}{BE}$. Let $ DA \cap BP = \{F\}$. $ \triangle ABC \sim \triangle BFC$. So $ AC \cdot FC = BC^2 = DC^2$ $ (..1..)$ Menelaus on $ \triangle ABF$, for $ P,E,C$ : $ \frac{AE}{BE} \frac{BF}{FP} \frac{FC}{AC}= 1$ $ (..2..)$ Menelaus on $ \triangle ABF$, for $ P,N,D$ : $ \frac{AN}{BN} \frac{BF}{FP} \frac{FD}{AD}= 1$ $ (..3..)$ $ \frac {(..2..)}{(..3..)} \Rightarrow \frac{AE}{BE} = \frac{AC}{FC} \frac{FD}{AD}$ $ (..4..)$. If you write $ (..1..)$ on $ (..4..)$, you will get $ \frac{AE}{BE} = \frac{AC}{DC}$. $ (..5..)$ The second part is finding $ \frac{BC}{TC}$. Menelaus on $ \triangle ABC$, for $ N,T,D$ : $ \frac{AN}{BN} \frac{BT}{CT} \frac{CD}{AD}= 1$ $ (..6..)$ So $ \frac{BC}{TC} = \frac{BT}{TC} + 1 = \frac{AD}{CD} + 1 = \frac{AC + DC}{DC} + 1 = \frac{AC}{DC} + 2$ $ (..7..)$ $ (..7..) - (..5..) \Rightarrow \frac{BC}{TC} - \frac{AE}{BE} = 2$

Find the BC/TC - EA/EB Turkey TST 2005 problem 2 Let N be midpoint of the side AB of a triangle ABC with m(A) greater than m(B). Let D be a point on the ray AC such that CD=BC and P be a point on the ray DN which lies on the same side of BC as A and satisfies the condition m(PBC)=m(A). The lines PC and AB intersect at E, and the lines BC and DP intersect at T. Determine the value of BC/TC - EA/EB. Sorry but what does m(A) mean? Is it <BAC?

yes, it is