Prove that $AD=BC$ if and only if $\measuredangle ADT\equiv \measuredangle TDC$. 
Let $ABC$ be a triangle, $AD$ one of it's heights and $G$ it's centroid. $DS$ is the bisector of $\measuredangle BDA$ with $S\in AB$, and $SG\cap AC=\{T\}$. Prove that $AD=BC$ if and only if $\measuredangle ADT\equiv \measuredangle TDC$.


Fist I considerd $AD=BC$ and realised that I had to prove that $DT$ is the bisector of $\measuredangle ADC$, so proving that $\frac{AD}{DC}=\frac{AT}{TC}$ would be enough. $\frac{AD}{DC}=\frac{AT}{TC}\iff\frac{BC}{DC}=\frac{AT}{TC}$. From $DS$ is the bisecor of$\measuredangle BDA$ $\implies$ $\frac{AD}{BD}=\frac{AS}{BS}\iff\frac{BC}{BD}=\frac{AS}{BS}$. I got a little stuck here. What shoud I do next or my abordation is wrong? ($M$ and $P$ are only to find the position of $G$) or maybe a vectorial approach is possible, but I don't see a solution that way.
 A: Let $DT$ and $DS$ meet parallel to $BC$ through $A$ at $E$ and $F$. Then we see that $AF = AD$ since $\angle FDA = 45^\circ$ and $\angle DAF = 90^\circ$. 

Lemma: $EF =2BC$
Proof: Since $\Delta BCG\sim \Delta EFG $ and $AG:GM =2:1$ we have also $EF: BC =2:1$

Now if $DE$ is angle bisector for $\angle ADC$ we have also $AE = AD$, so $EF = 2AD$. So, by lemma we have $AD = BC$.   
Vice versa: if $BC =AD$ then $AF =BC$. Since again $EF = 2BC$ so by lemma we have $AE=BC$ and thus $DT$ is angle bisector for $\angle ADC$.

A: Reformulation of Problem Statement
Let $ABC$ be a triangle and $D$ the foot of the altitude from $A$.  The two angular bisectors $\ell_b$ and $\ell_c$ of the right angle $\angle ADB$ are defined as follows:


*

*If $D$ lies between $B$ and $C$, then $\ell_b$ and $\ell_c$ respectively intersect $AB$ and $AC$ internally;

*If $B$ lies between $C$ and $D$, then $\ell_b$ intersects $AB$ externally, whilst $\ell_c$ intersects $AC$ internally;

*If $C$ lies between $B$ and $D$, then $\ell_b$ intersects $AB$ internally, whereas $\ell_c$ intersects $AC$ externally.


Suppose that $\ell_b$ meets $AB$ at $S$ (if $B=D$, then $S$ is defined to be $D$), and $\ell_c$ meets $AC$ at $T$ (if $C=D$, then $T$ is defined to be $D$).  Then, $G$ lies on the straight line $ST$ if and only if $|AD|=|BC|$.
Compilation of Proofs from the Comments
Solution I: Analytic Geometry
Let $A:=(0,a)$, $B:=(b,0)$, $C:=(c,0)$, and $D:=(0,0)$, where $a$, $b$, and $c$ are real numbers such that $a>0$ and $b<c$.  Note that $\ell_b$ is given by the equation
$$x+y=0\,,$$
whilst $\ell_c$ is given by the equation
$$x-y=0\,.$$
The line $AB$ is given by the equation
$$a(x-b)+by=0\,,$$
whereas the line $AC$ is given by the equation
$$a(x-c)+cy=0\,.$$
Therefore,
$$S=\left(\frac{ab}{a-b},\frac{ab}{b-a}\right)$$
while
$$T=\left(\frac{ac}{a+c},\frac{ac}{a+c}\right)\,.$$
(If $a-b=0$, then $S$ is the point at infinity on the line $\ell_b$.  If $a+c=0$, then $T$ is the point at infinity on the line $\ell_c$.)
Consequently, the line $ST$ is given by the equation
$$a(b+c)x+(2bc+ab-ac)y-2abc=0\,.$$
(This equation for $ST$ is good even when $a-b=0$ or when $a+c=0$.)  The centroid $$G=\dfrac{A+B+C}{3}=\left(\dfrac{b+c}{3},\dfrac{a}{3}\right)$$ is on $ST$ if and only if
$$a(b-c)(a+b-c)=a(b+c)^2+(2bc+ab-ac)a-6abc=0\,,$$
or equivalently, $|AD|=a=c-b=|BC|$.
Solution II: Menelaus's Theorem
Draw $SG$ to meet the line $AC$ at $T'$, and the line $BC$ at $X$.  Let $M$ be the midpoint of $BC$.  In what follows, the distances are measured with signs, meaning that $PQ=-QP$ for any two points $P$ and $Q$.
By Menelaus's Theorem on the triangle $ABM$ with the collinear points $S\in AB$, $G\in MA$, and $X\in BM$, we have
$$\frac{BS}{SA}\cdot \frac{AG}{GM}\cdot \frac{MX}{XB}=-1\,.$$
Since $\dfrac{AG}{GM}=2$, we obtain
$$\frac{BS}{SA}=-\frac{XB}{2\,MX}=\frac{BX}{2\,MX}\,.$$
By Menelaus's Theorem on the triangle $ACM$ with the collinear points $T'\in AC$, $G\in MA$, and $X\in CM$, we have
$$\frac{CT'}{T'A}\cdot \frac{AG}{GM}\cdot \frac{MX}{XC}=-1\,.$$
Since $\dfrac{AG}{GM}=2$, we obtain
$$\frac{CT'}{T'A}=-\frac{XC}{2\,MX}=\frac{CX}{2\,MX}\,.$$
Consequently,
$$\frac{BS}{SA}+\frac{CT'}{T'A}=\frac{BX}{2\,MX}+\frac{CX}{2\,MX}=\frac{BX+CX}{2\,MX}\,.$$
Since
$$BX=BM+MX\,,\,\,CX=CM+MX\,,\text{ and }BM+CM=0\,,$$ we conclude that $BX+CX=2\,MX$, whence
$$\frac{BS}{SA}+\frac{CT'}{T'A}=1\,.$$
Therefore, $G$ lies on $ST$ if and only if $T=T'$.  From the previous paragraph, this is equivalent to $\dfrac{BS}{SA}+\dfrac{CT}{TA}=1$.  By the Angular Bisector Theorem,
$$\dfrac{|BS|}{|SA|}=\frac{|BD|}{|AD|}\text{ and }\dfrac{|CT|}{|TA|}=\frac{|CD|}{|AD|}\,.$$
If $BC$ is chosen to be positive, making $CB$ negative, then we can write
$$\frac{BS}{SA}=\frac{BD}{|AD|}\text{ and }\frac{CT}{TA}=\frac{DC}{|AD|}\,.$$
Hence,
$$\dfrac{BS}{SA}+\dfrac{CT}{TA}=\frac{BD}{|AD|}+\frac{DC}{|AD|}=\frac{BC}{|AD|}=\frac{|BC|}{|AD|}\,.$$
Consequently, $\dfrac{BS}{SA}+\dfrac{CT}{TA}=1$ if and only if $|AD|=|BC|$, establishing the result.
P.S.  From the second solution, the following is true.  Let $ABC$ be a triangle and $N$ a point on the line $BC$.  Suppose that $U$ is a point on the line $AN$.  For points $V$ and $W$ on the line $AB$ and the line $AC$, respectively, $U$ lies on the line $VW$ if and only if
$$\frac{NC}{BC}\cdot \frac{BV}{VA}+\frac{BN}{BC}\cdot \frac{CW}{WA}=\frac{NU}{UA}\,.$$
(The distances here are also measured with signs.)
