For reference: In the triangle $ABC$ ($\angle B = 90^o$) through the point $N$ exterior of this triangle is drawn $NP$ perndicular to $BC$ ($P$ is the midpoint of $BC$). Calculate $AN$; if $AB = 3\sqrt29$ and $BC=4\sqrt29$.(Answer:$29$)
My progress:
Draw $CD$ and $BD$ $\implies \triangle BCD$ (right)
$PC = PD =2\sqrt29$
$\triangle FPD \sim \triangle FBA \implies$
$\frac{FP}{BF}=\frac{FA}{FD}=\frac{3\sqrt29}{3\sqrt29} =\frac{3}{2} $
$\frac{3BF}{2}+BF = 2\sqrt29 \implies BF = \frac{4\sqrt29}{5}\\ \therefore FP = \frac{6\sqrt29}{5}$
$\triangle ABF: (3\sqrt29)^2+(\frac{4\sqrt29}{5})^2 = AF^2\implies AF=\frac{\sqrt{6989}}{5}$
??? I don't think this is the way to go
There is a problem in the question text so it will be considered that $CN \perp NB$
By Math's hint:
$If ∠BNC=90^∘ \rightarrow PN=BP=\frac{B}{2}$
Extend AB and drop a perp from N. Say it meets AB at H then right triangle △AHN. – $\triangle AHN: (AB+PN)^2+BP^2=AN^2\implies \\ (3\sqrt29+2\sqrt29)^2+(2\sqrt29)^2 = AN^2\\ 25\cdot29+4\cdot29=AH^2 \therefore \boxed{AH = 29}$