Triangle in Triangle I have the lengths of three sides of an acute triangle ABC as shown below. Assume a point P on the side AB such that, if Q is the projection of P onto BC, R is the projection of Q onto CA, P becomes the projection of R onto AB. How can I Find the length PB.


Here is another approach, which works directly and systematically. Call the angles of the triangle $A, B, C$ and the sides opposite those angles $a, b, c$. We use the cosine rule $a^2=b^2+c^2-2bc \cos A$ so that $\cos A=\cfrac {a^2+b^2+c^2-2a^2}{2bc}$ and if I define $2d^2=a^2+b^2+c^2$ then $\cos A=\cfrac {d^2-a^2}{bc}, \cos B=\cfrac {d^2-b^2}{ac}, \cos C=\cfrac {d^2-c^2}{ab}$. 
Let BP=p.
Then $$p=c-AP=c-AR\cos A= c-(b-CR)\cos A=c-b\cos A+CR\cos A=$$
$$=c-b\cos A+CQ\cos C\cos A=c-b\cos A+(a-BQ)\cos C\cos A=$$
$$=c-b\cos A+a\cos A\cos C-BQ\cos A\cos C=$$
$$=c-b\cos A+a\cos A\cos C-p\cos A\cos B\cos C$$
So that $$p\left(1+\cos A\cos B\cos C\right)=c-b\cos A+a\cos A\cos C=$$$$=c-\frac {d^2-a^2}c+\frac {(d^2-a^2)(d^2-c^2)}{b^2c}=\frac {b^2c^2-b^2d^2+b^2a^2+d^4-a^2d^2-c^2d^2+a^2c^2}{b^2c}$$
Now with $a^2+b^2+c^2=2d^2$ this is equal to $$\frac {a^2b^2+b^2c^2+a^2c^2-d^4}{b^2c}$$
We also have $$1+\cos A\cos B\cos C=\frac {a^2b^2c^2+d^6-(a^2+b^2+c^2)d^4+(a^2b^2+b^2c^2+a^2c^2)d^2-a^2b^2c^2}{a^2b^2c^2}=\frac {d^2(a^2b^2+b^2c^2+c^2a^2-d^4)}{a^2b^2c^2}$$
So that everything cancels nicely to give $$p=\frac {a^2c}{d^2}=\frac {2a^2c}{a^2+b^2+c^2}$$
