Let $ X,Y$ be the orthogonal projections of $ D$ on $ AB,AC$ and $ B'$ the reflection of $ B$ about $ AD.$ Then it follows that $ B'$ lies on the circle centered at $ D$ that goes through $ B,C,$ due to $ \angle BB'C = 90^{\circ} + \frac {_1}{^2}\angle A.$ Hence, $ DY$ is perpendicular bisector of $ B'C.$ WLOG assume that $ AC > AB$ $ \Longrightarrow$ $ B'Y = CY = \frac {_1}{^2}(AC - AB).$ Therefore $ AX = AY = AB' + B'Y = AB + \frac {_1}{^2}(AC - AB) = \frac {_1}{^2}(AB + AC) \ \ (1)$ On the other hand, by Casey's theorem for $ (A,B,C,\omega),$ all tangent to $(O),$ we have $ AT \cdot BC = AC \cdot BM + AB \cdot CM = \frac {_1}{^2}BC(AB + AC)$ $ \Longrightarrow AP = AT = \frac {_1}{^2}(AB + AC) \ \ (2)$ From $ (1)$ and $ (2),$ we conclude that $ X$ and $ P$ are identical $ \Longrightarrow \angle APD = 90^{\circ}.$

Let $ M$ be midpoint of $ BC$ and $ D_1,D_2$ respectively be the projections of $ D$ onto $ AB$, $ AC$. Then by Simpson line theorem, we have $ D_1,M,D_2$ are collinear. Let $ X\equiv AD\cap D_1D_2$. In the other hand, since $ AD$ is the internal bisector wrt $ \angle BAC$ of $ \triangle ABC$. Therefore, $ AD\perp D_1D_2$ at $ X$, which implies $ \angle MXD = 90^{\circ}$. Thus $ X\in (W)$. Moreover, we have $ AD_1^2 = \overline {AX}.\overline {AD} = \mathcal {P}_{A/(W)} = AT^2$ $ \Longrightarrow$ $ AD_1 = AT$, which implies that $ D_1\equiv P$, which leads to the result of the problem. $ \square$

Let $ M$ be the midpoint of $ BC$. $ \frac{AD}{CD}=\frac{AT}{CM} \Rightarrow \frac{AD}{AP}=\frac{CD}{CM}$ and $ \angle MCD=\angle DAB \Rightarrow \angle APD=\angle CMD=90^o$

shoki wrote: 1-Let $ \triangle ABC$ be a triangle and $ (O)$ its circumcircle. $ D$ is the midpoint of arc $ BC$ which doesn't contain $ A$. We draw a circle $ W$ that is tangent internally to $ (O)$ at $ D$ and tangent to $ BC$.We draw the tangent $ AT$ from $ A$ to circle $ W$.$ P$ is taken on $ AB$ such that $ AP = AT$.$ P$ and $ T$ are at the same side wrt $ A$.PROVE $ \angle APD = 90^\circ$. Complete trig bashing! Let $ DP' \perp AB$. Then it is enough to prove that $ AP'=AP=AT$. We have $ AP'=AD \cos \frac B2$. Sine law in $ \triangle ABD$ gives,\begin{align*} &AD=2R \sin (B+A/2) \\ \Longrightarrow & AP'=R\cdot 2\sin (B+A/2) \cos \frac B2=R(\sin C+\sin B)=\frac{b+c}{2}\end{align*} Let $ M$ be the midpoint of $ BC$. Now from power of point, $ AT^2=AW^2-MW^2$ From Cosine law in $ \triangle AMD$, \begin{align*} & AW^2=AM^2+MW^2-2AM\cdot MW\cos \angle AMD \\ \iff & AT^2=AW^2-MW^2=AM^2+2AM\cdot MW\cos \angle AMH'\end{align*} Here $ AH' \perp MO$. We can see that $ H'M=AH=AM \cos \angle AMH'$ We know that $ AM^2=\frac{2b^2+2c^2-a^2}{4}=\frac{b^2+c^2}{4}+\frac{b^2+c^2-a^2}{4}=\frac{b^2+c^2}{4}+\frac{bc \cos A}{2}$. Also, $ AH=b \sin C$, and $ MD=\frac a2 \tan \frac A2$ So, \begin{align*} 2AM\cdot MW\cos \angle AMH'= & AH\cdot MD \\ = & \frac 12 ab \sin C \tan \frac A2 \\ = & \frac{bc}{2} \sin A \frac{\sin \frac A2}{\cos \frac A2} \\ = & \frac{bc}{2}\cdot 2 \sin^2 \frac A2 \end{align*} At last, \begin{align*} AP^2= & \frac{b^2+c^2}{4}+\frac{bc \cos A}{2}+\frac{bc}{2}\cdot 2 \sin^2 \frac A2 \\ = & \frac 14 \left (b^2+c^2+2bc (\cos A+2\sin^2 A/2)\right ) \\ = & \frac{(b+c)^2}{4}\end{align*} So, $ AP=AP'=\frac{b+c}{2}$, and we are done!

livetolove212 wrote: $ \frac {AD}{CD} = \frac {AT}{CM}$ Why?

Let $M$ be midpoint of $BC$ and $BC \cap AD = X$ $W$ is clearly tangent to $BC$ at $M$. $AD \cap W = K$ $AT^2 = AK.AD = AP^2 \Longrightarrow PK \perp AD \Longrightarrow P,K,M$ are collinear $\Longrightarrow \angle PBD = \angle \frac {A} {2}+ \angle C = \angle AXB = \angle DXM = \angle PMD$ So $P,B,M,C$ are concyclic and $ \angle BPD = 180 - \angle BMD = 90$