In a triangle ABC, D is …

It’s given that the point DD lies on the side BCBC. Consider the side BCBC and angle CC in the diagram. Also, the circle is centered at AA, as we need to keep AD=ACAD=AC. We can see that, the angle CC has to be acute, if a point DD with AD=ACAD=AC, has to lie on the side BCBC. Once, it is confirmed that the third one is the only possible scenario, we can move ahead with the following solution. A simple construction helps solve the problem. Drop a perpendicular from AA on the side BCBC. Let it meet the side at point PP. From the diagram, we can use different approaches to prove that AB>ADAB>AD BP>DPBP>DP and Pythagoras’ theorem for triangles APDAPD and APBAPB, with common side APAP Trigonometry : We can write expressions for sinθ1sinθ1 and sinθ2sinθ2 and compare. Here, θ1>θ2θ1>θ2 because of exterior angle property (triangle ADBADB) We can also simply state that, the farther we are from the point PP, the distance is clearly increasing, hence AB>ADAB>AD. But this is same as the 11st point. Hoping this was helpful. Let me know if there’s an issue.