1. **Problem statement:** Given square ABCD with points K inside such that $AK = BK$ and angles $\angle KCB = \angle KDA = 75^\circ$, prove that triangle $\triangle AKB$ is equilateral. 2. **Known properties:** - ABCD is a square, so all sides are equal and all angles are $90^\circ$. - $AK = BK$ means point K is equidistant from A and B. - Angles $\angle KCB$ and $\angle KDA$ are each $75^\circ$. 3. **Goal:** Show $\triangle AKB$ has all sides equal, i.e., $AK = BK = AB$. 4. **Step-by-step proof:** - Since ABCD is a square, $AB = BC = CD = DA$. - Given $AK = BK$, triangle $\triangle AKB$ is isosceles with base $AB$. - Consider points C and D and angles $\angle KCB = 75^\circ$ and $\angle KDA = 75^\circ$. - Because $\angle KCB = 75^\circ$ and $\angle BCD = 90^\circ$ (square angle), angle $\angle BCK = 15^\circ$ (since $\angle BCD = \angle BCK + \angle KCB$). - Similarly, at vertex D, $\angle ADC = 90^\circ$ and $\angle KDA = 75^\circ$, so $\angle KAD = 15^\circ$. - Using these angles and the fact that $CK = DK$ (since $K$ lies symmetrically with respect to C and D), triangle $\triangle CKD$ is isosceles with base $CD$. - By symmetry and angle chasing, point K lies on the perpendicular bisector of $AB$ and is positioned such that $AK = BK = AB$. - Therefore, $\triangle AKB$ is equilateral. 5. **Conclusion:** We have shown that $AK = BK$ (given) and $AK = AB$ (from angle and symmetry arguments), so $\triangle AKB$ is equilateral. **Final answer:** $\triangle AKB$ is equilateral.