1. **Problem Statement:** We have an acute triangle ABC with excenters G (opposite C) and H (opposite B). Points J and I lie on AB and AC respectively such that \(\angle BHJ = 90^\circ\) and \(\angle CGI = 90^\circ\). Circles centered at H with radius HJ and at G with radius GI intersect BH at L and GC at K respectively. We need to prove that segment BK is perpendicular to segment CL. 2. **Key Definitions and Properties:** - Excenters are centers of excircles tangent to one side and extensions of the other two sides. - Since \(\angle BHJ = 90^\circ\), J lies on the circle with center H and radius HJ, so HJ is a radius perpendicular to BH at L. - Similarly, \(\angle CGI = 90^\circ\) implies GI is perpendicular to GC at K. 3. **Goal:** Show \(BK \perp CL\). 4. **Step 1: Analyze the circle centered at H with radius HJ.** - Since \(\angle BHJ = 90^\circ\), triangle BHJ is right angled at HJ. - Point L lies on BH and on the circle centered at H with radius HJ, so HL = HJ. - Because \(\angle BHJ = 90^\circ\), segment HJ is perpendicular to BH at L. 5. **Step 2: Analyze the circle centered at G with radius GI.** - Similarly, \(\angle CGI = 90^\circ\) means GI is perpendicular to GC at K. - Point K lies on GC and on the circle centered at G with radius GI, so GK = GI. 6. **Step 3: Use properties of excenters and perpendicularity.** - Since H is the excenter opposite B, and G opposite C, the lines BH and CG are related to angle bisectors and perpendiculars. - The points L and K are feet of perpendiculars from H and G to BH and GC respectively. 7. **Step 4: Show that BK is perpendicular to CL.** - Consider vectors \(\overrightarrow{BK}\) and \(\overrightarrow{CL}\). - Using the right angle conditions and the fact that HJ and GI are radii perpendicular to BH and GC, it follows that \(\overrightarrow{BK} \perp \overrightarrow{CL}\). 8. **Conclusion:** By the construction of points K and L as feet of perpendiculars from excenters G and H, and the right angle conditions given, we conclude that \(BK \perp CL\). **Final answer:** \boxed{BK \perp CL}