Hint: A_1, B_1, C_1 the orthogonal projection of D onto BC, CA, AB express DB_1 in terms of DA, DC, R the circumradius of ABC DA_1 in terms of DB, DC DC_1 in terms of DA, DB

Draw the perpendiculars DX=x,DY=y,DZ=z from D to CB,BA,AC. Then:x+y=z...(1) (Some time ago I have posted this problem in: http://www.personal.us.es/rbarroso/trianguloscabri/pro150.htm ) Sorry for the spanish. Also by similar pairwise triangles: BYD,CZD: \frac{y}{BD}=\frac{z}{CD} ..(2) BXD,AZD: \frac{x}{BD}=\frac{z}{AD} ..(3) From (1),(2) and (3): \frac{1}{BD}=\frac{1}{AD}+\frac{1}{CD} QED.

I can see that a solution has been posted, but I took the time to write this, so.. At first I struggled really hard to find a non-computational proof, but I failed. However, the computational proof I found is far from nasty. Let $T=AD\cap BC$, and let $a,b,c=BC,CA,AB$ respectively. We use three things: $(1)$: triangles $TBD,TAC$ are similar, so $\frac {TD}{BD}=\frac{TC}b$; $(2)$: triangles $TDC,TBA$ are similar, so $\frac{TD}{CD}=\frac{TB}c$; $(3)$ from the transversal theorem we get $\frac {BN}{NA}TC=\frac{CM}{MA}TB+\frac{TD}{AD}BC\iff \frac ab TC=\frac ac TB+\frac{TD}{AD}a$. we deivde everything by $a$ and we get $\frac {TC}b=\frac{TB}c+\frac{TD}{AD}$. From $(1),(3)$ we get $\frac {TB}c=TD(\frac 1{BD}-\frac 1{AD})\ (*)$, and from $(*),(2)$ we get $\frac{TD}{CD}=TD(\frac 1{BD}-\frac 1{AD})$, and by dividing this last equality with $TD$ we get exactly what we want. I think it has been posted before, btw, because I remember struggling to find an inverseive proof some time ago. This idea was inspired by the fact that after an inversion of pole $D$ we would have to show that $DB'=DA'+DC'$, where, in general, $X'$ is the image of $X$ under the inversion, but this seems much harder and I could make no progress whatsoever.