Given a triangle
![ABC](/media/m/a/c/7/ac75dca5ddb22ad70f492e2e0a153f95.png)
, let
![D](/media/m/7/0/0/7006c4b57335ab717f8f20960577a9ef.png)
and
![E](/media/m/8/b/0/8b01e755d2253cb9a52f9e451d89ec11.png)
be points on the side
![BC](/media/m/5/0/0/5005d4d5eac1b420fbabb76c83fc63ad.png)
such that
![\angle BAD = \angle CAE](/media/m/1/2/d/12d0c02e90b4f86441a53564d254bec7.png)
. If
![M](/media/m/f/7/f/f7f312cf6ba459a332de8db3b8f906c4.png)
and
![N](/media/m/f/1/9/f19700f291b1f2255b011c11d686a4cd.png)
are, respectively, the points of tangency of the incircles of the triangles
![ABD](/media/m/a/5/4/a548bc577543629d304ecba1a042f910.png)
and
![ACE](/media/m/0/8/3/08350444735a91bb62351789317e7dac.png)
with the line
![BC](/media/m/5/0/0/5005d4d5eac1b420fbabb76c83fc63ad.png)
, then show that
%V0
Given a triangle $ABC$, let $D$ and $E$ be points on the side $BC$ such that $\angle BAD = \angle CAE$. If $M$ and $N$ are, respectively, the points of tangency of the incircles of the triangles $ABD$ and $ACE$ with the line $BC$, then show that
$$\frac{1}{MB}+\frac{1}{MD}= \frac{1}{NC}+\frac{1}{NE}.$$