In the figure, $\Delta ABC$ is right-angled at $B$. $I$ is the incentre. $KI \perp CE $ and $JI \perp AI$. If $EK=DJ$, show that $AB=BC.$ 
In the figure, $\Delta ABC$ is right-angled at $B$. $I$ is the incentre. $KI \perp CE $ and $JI \perp AI$. If $EK=DJ$, show that $AB=BC.$
 A: There are various ways you can approach this problem. Here is one straight-forward argument. It is not the only one though. 
Lines $AI$ and $CI$ are angle bisectors of the angles at vertices $A$ and $C$ respectively. Then $$\angle \, AKI = \angle \, CIK + \angle \, ICK = 90^{\circ} + \frac{1}{2} \, \angle \, C$$ Also $$\angle \, AEI = \angle \, CBE + \angle \, BCE = 90^{\circ} + \frac{1}{2} \, \angle \, C  = \angle \, AKI$$ The latter fact, combined with the observation that $ \angle \, AIK = \angle \, IAE = \frac{1}{2} \, \angle A$ implies that triangles  $AIK$ and $AIE$ are congruent and thus $AK = AE$ as well as $IK=IE$. triangles  $AIK$ and $AIE$ are congruent and thus $AK = AE$ as well as $IK=IE$. Analogously, triangles $CIJ$ and $CID$ are congruent and consequently $CJ = CD$ as well as $IJ=ID$. It follows from all of these arguments, that lines $AI$ and $CI$ are orthogonal bisector of segments $EK$ and $JJ'$ ($J' \in AB$ is the symmetric of $J$ with respect to $AI$) and $CI$ is also orthogonal bisector of segments $DJ$ and $KK'$ ($K' \in AB$ is the symmetric of $K$ with respect to $CI$).
Triangles $EIK$ and $DIJ$ are isosceles with right angles at vertex $I$. If $EK = DJ$ then $EIK$ and $DIJ$ are congruent and 
$$EI = KI = JI = DI = K'I = J'I$$ so they lie on a common circle $c_I$ centered at $I$ and of radius $DI$, passing through the points $E, \, K, \, J, \, D, \, K', \,  J'$.  Also, the three triangles $JKI, \, DK'I, \, DJ'I$ are all congruent because they are isosceles and the angles at their common vertex $I$ are all equal to $45^{\circ}$. Hence $JK = DK' = RJ'$. 
Moreover, triangle $AKI$ is congruent to $CJI$ because $JI = KI$, together with $\angle \, CIJ = \angle \, AIK = 45^{\circ}$ and $\angle \,  IJC = \angle \,IKA $. Hence $AK = CJ$ and so
$$AE = AK = CJ = DC$$
Now, you can show that triangles $BK'I$ and $BJ'I$ are also congruent, and so $J'B = BK'$. Putting all together,
$$AB = AE + EJ' + J'B = DC + K'D + BK ' = BC$$
Edit. Alternatively, as pointed out by tartan,  the congruence between triangles $AKI$ and $CJI$ implies that $\angle \, A = 2 \, \angle \, IAK = 2 \, \angle \, ICJ = \angle \, C$ which means that $ABC$ is isosceles and $AB = BC$.
