please anyone?

Let $\alpha = \angle PAB$, $\beta = \angle PBC$ and $\gamma = \angle PCA$, with $\alpha + \beta + \gamma = 90^{\circ}$. By trig Ceva we have $$ \tan(\alpha)\cdot\frac{\sin(\beta )}{\sin(45^{\circ}-\beta )}\cdot\frac{\sin(\gamma )}{\sin(45^{\circ}-\gamma )}=1\iff$$$$\iff 2\cot(\beta + \gamma)\cdot\frac{\tan(\beta )}{1-\tan(\beta )}\cdot \frac{\tan(\gamma )}{1-\tan(\gamma )} =1\iff$$$$\iff \frac{2xy(1-xy)}{(1-x)(1-y)(x+y)} =1,$$where we let $x=\tan(\beta )$ and $y=\tan(\gamma )$. After some manipulation, the latter is equivalent to $$\frac{xy+x+y-1}{(1-x)(1-y)} =\frac{xy+x+y-1}{1-xy}.$$Now there are two cases: if $xy+x+y-1=0$, or $(1-x)(1-y)=1-xy$. Observe that the first case is equivalent to $\tan(\beta + \gamma) = \frac{1-xy}{x+y} = 1$, so $\alpha = 45^{\circ}$, exactly what we wanted to prove. In the second case, the equality reduces to $(2x-1)(2y-1)=1$, which is false since $\beta, \gamma \in\left(0^{\circ},45^{\circ}\right)\iff x,y\in (0,1)\iff 2x-1,2y-1\in (-1,1)$.

Let $D$ be the point on $BC$ with $AD\perp BC$. If $AP$ is not perpendicular to $BC$, without loss of generality, assume $P$ is inside the triangle $ABD$. Write $\widehat{PBC}=x+y$ and $\widehat{PCB}=y$. From the given angle equality, it is easy to see that $\widehat{PAD}=x$. Let $F$ be the point on $AD$ with $|PA|=|PF|$. Firstly, we have $\widehat{PFD}=x$. Then, when we consider the triangle $PBC$, we have $$|FB|=|FC|\,\text{ and }\,\widehat{PFC}-\widehat{PFB}=2x=2\cdot (\widehat{PBC}-\widehat{PCB})$$ This implies $F$ is the circumcenter of the triangle $PBC$, and hence we get $|FB|=|FP|$. On the other hand, let $K$ be the point on $AB$ such that $PK\perp AD$. Since $\widehat{AKF}=90^\circ$ and $KP\perp AF$, we get $|FB|>|FK|>|FP|$, which leads a contradiction. As a result, we find $AP\perp BC$.

see the figure:

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Use trig ceva and you will get 2 cases,one of them leading to $\angle{BAP} = \angle{45} $ which implies $AP \perp BC$ because $ABC$ is isosceles.

Tsikaloudakis wrote: see the figure: Why is AB and BZ same?

changeΑΒ : with ΑΡ.