The PDF Solution is attached with my post. I don't know latex, so I used word math.

Attachments:
Equal inradii Solution.pdf (447kb)

Observe that $r_{ABD}=r_{ACD}=\frac{BD.AB}{AB+AD+BD}=\frac{CD.AB}{AC+AD+CD}$ $\implies \frac{AB+AD}{BD}=\frac{AC+AD}{CD}$. So, by Law of Sines, we get $\tan{\phi}+\sec{\phi} = \frac{\sin{\phi}+\sin{C}}{\sin{(\phi -C)}}$. Solving this ugly thing, we would get the desired $\tan^2{\phi/2}=\tan{C/2}$

Some results first. 1: In a triangle $ ABC $ with the usual notations, $ \tan \frac{A}{2} = \sqrt{\frac{(s-b)(s-c)}{s(s-a)}} $. Further, if it is right angled at $ B $, then $ \tan \frac{A}{2}= \sqrt{ \frac{b-c}{b+c}} $. 2: The Equal Incircles Theorem: Suppose that points $ D,E $ on the side $ BC $ of a $ \triangle ABC $ are such that the incircles of $ \triangle ABD, \triangle ACE $ are congruent. Then the incircles of $ \triangle ABE, \triangle ACD $ are also congruent. Back to the the problem. Let the reflection of $ D $ in $ AB $ be $ E $. Then the triangles $ ABE,ABD $ are congruent, hence the three triangles $ AEB, ABD, ADC $ have congruent incircles. Let their radii be $ r $. By 2 the incircles of $ \triangle AED, \triangle ABC $ are also congruent. Let their radii be $ \rho $. We then have, \[ \rho \cdot s_{\triangle AED}= [AED]= [AEB] + [ABD]= r(s_{\triangle ABE}+s_{\triangle ADB}) \] \[ \implies \frac{\rho}{r} = \frac{s_{\triangle AEB}+s_{\triangle ADB}}{s_{\triangle AED}}=\frac{s_{\triangle AED}+AB}{s_{\triangle AED}} = 1 + \frac{AB}{s_{\triangle AED}} \] Similarly, we also have \[ \frac{\rho}{r}= 1+ \frac{AD}{s_{\triangle ABC}} \] Equating both, we get \[ \frac{AB}{AD}=\frac{s_{\triangle AED}}{s_{\triangle ABC}} = \frac{[AED]}{[ABC]}=\frac{ED}{BC}=\frac{2BD}{BC} (\star) \] The relation $ \tan^2 \frac{\phi}{2} = \tan \frac{C}{2} $ is equivalent to \[ \frac{AD-BD}{AD+BD}=\sqrt{\frac{AC-BC}{AC+BC}} \] \[ \iff \frac{AD^2+BD^2}{2AD.BD}=\frac{AC}{BC} \] \[ \iff AD^2+BD^2=AC.AB \iff AD^2=\frac{AB(AC+AB)}{2} \] But from $ (\star) $, $ AD^2 $ is a root of the equation $ 4x^4-4x^2 \cdot AB^2-AB^2 \cdot BC^2=0 $, therefore, \[ AD^2=\frac{4AB^2+\sqrt{16AB^4+16AB^2BC^2}}{8}=\frac{AB(AC+AB)}{2} \blacksquare \]

Dear Mathlinkers, to be more geometric, can we precise D wrt ABC? Sincerely Jean-Louis