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In $\triangle ABC$, $\angle A=2\angle C$. $(O)$ is its $A-$excircle, and, $M$ is the mid-point of $AC$. $OM\cap BC=D$. Show that,
$[1]$ $AD$ bisects $\angle OAC$
$[2]$ $BA=BD$

sowmitra wrote:. $(O)$ is its $A-$excircle

I don't understand this line.Please clarify it.

$O$ is the centre of the excircle opposite to $A$ , to match with notations used in books , $O=I_A$

[1]
Let $AD\cap OC=N$
For triangle $\Delta AOC$ ;
$\frac{AP}{PO}\cdot \frac{ON}{NC}\cdot \frac{CM}{MA}=1$ [ceva]
or, $\frac{AP}{PO}=\frac{NC}{NO}$

Now let $I$ be the incentre of triangle $\Delta ABC$ ...... $\therefore I\epsilon AP$
$(A,I,P,O) $ harmonic
$\therefore \frac{AI}{IP}=\frac{AO}{PO}$
or,$\frac{AO}{PO}=\frac{AC}{CP}$ [angle bisector theorem]
or,$\frac{AO}{PO}=\frac{AC}{AP}$ [$CP=AP$ because $\angle A=2\angle C$]
or,$\frac{AP}{PO}=\frac{AC}{AO}$

$\therefore \frac{NC}{NO}=\frac{AC}{AO}$ [$=\frac{AP}{PO}$]
So, from the converse of angle bisector theorem ; $AD$ bisects $\angle OAC$


[2]
$\angle BAD=\angle BAP\angle PAD=\angle C+\frac{\angle C}{2}$
$\angle BDA=\angle DCA+\angle DAC=\angle C+\frac{\angle C}{2}$
$\angle BAD=\angle BDA$
So, $BA=BD$

photon wrote:$O$ is the centre of the excircle opposite to $A$ , to match with notations used in books , $O=I_A$

Thanks for clarify it.BTW,here is a very easy solution of part $1$:
Let $AO\cap BC=U$.Join $O,C$.By $\text{Menelaus's theorem}$ we get,$\frac{AO}{OU}\times \frac{UD}{DC}\times \frac{CM}{MA}=1$.Or,$\frac{AC}{CU}\times \frac{UD}{DC}=1$.Or,$\frac{UD}{DC}$=$\frac{CU}{AC}$.
Now,$AU=UC$.$\therefore$ $\frac{UD}{DC}$=$\frac{AU}{AC}$.
By the $\text{Converse of Angle Bisector Theorem}$,$AD$ bisects $\angle OAC$.

The second part is same as Tahmid vaia's proof.