help please

Draw ID$\perp$AC. Then ID = r, the inradius of $\Delta$ABC. Observe EF$\parallel$BC and hence $\angle$AEF = $\angle$ABC = $90^\circ$. Hence $\angle$AIF =$90^\circ$. Therefore $ID^{2}$ = F D · DA. If a>c, then F A > DA and we have DA = s − a, and F D = F A − DA =$\frac{b}{2}$ − (s − a). Thus we obtain : $r^{2}$=$\frac{(b+c-a)(a-c)}{4}$ But r = (c + a − b)/2. Thus we obtain $(c+a-b)^{2}$ = (b + c − a)(a − c). Simplification gives 3b = 3a + c. Squaring both sides and using $b^{2}=a^{2}+c^{2}$, We obtain:4c = 3a. Hence BC/BA = a/c = $\boxed{4/3}$.

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I solved this question by drawing the perpendicular bisector of segment EF

then please send the solution

naren28 wrote: Let ABC be a right triangle with $\angle B = 90^{\circ}$.Let E and F be respectively the midpoints of AB and AC.Suppose the incentre I of ABC lies on the circumcircle of triangle AEF,find the ratio BC/AB. Construct $I^*$ as incenter of $\triangle AFE$ then observe $\triangle FI^*I$ is isosceles right $\triangle$ with $FI=II^*=EI=x$ and also $AI=2x$ then Pythagoras gives $AF=\sqrt{5}x$ and also let $AI\cap FE=K$ now as $\triangle AFK\sim \triangle AIE \implies \frac{AF}{FK}=\frac{AI}{IE}=2$ now By angle bisector Theorem we get $AE=2KE$ so by pythagoras we get $FE=\frac{8x}{\sqrt{5}}, AE=\frac{6x}{\sqrt{5}}$ Hence $\frac{BC}{AB}=\frac{4}{3}$ $\blacksquare$

Just draw the inradii $IX,IY$ on $BC and AB$.by a bit of angle chasing Inradii=$BC/4$.Prove that $IY$ is tangent on the circumcircle of $AEF$.Then use power of point $IY^2=YE×YA$.use $YE=BE-BY=AB/2-BC/4$.Solve the equation and we are done. For the people who had done the first chapter of egmo,it will be like a piece of cake

Drop a perpendicular from $I$ to $AC$ and let the touch-point with $AC$ be $G$, it is easy to check $\angle AEF=\angle AIF=\angle ABC=90^{\circ}$, now we length chase as we have $r^2=FG\cdot AG$, we have \begin{align*} AG&=\frac{AB-BC+CA}{2}\\ FG&=\frac{AC}{2}-\left(\frac{AB-BC+CA}{2}\right)=\frac{BC-AB}{2}\\ \implies r^2&=\frac{(AB-BC+CA)(BC-AB)}{4} \end{align*}and $r=\frac{AB+BC-CA}{2}$ so we have $$(AB+BC-CA)^2=(BC-AB)(AC+AB-BC)\iff 3BC+AB-3AC=0$$substitute PGT, then do squaring to get $\boxed{\frac{BC}{AB}=\frac{4}{3}}$.