Pause and Solve: A single ROTATION solves the problem.

Given a triangle with ABC, erect squares ADEB and ACFG outwardly on the sides of triangle ABC. Let M be the midpoint of BC. Show that AM is perpendicular to GD.
A general tip for IOQM, RMO, INMO, IMO, AIME, AMC: When you see a bunch of squares around a common point, Rotate by 90 degrees about that point!
Edited by: Shreesha Shenoy Badthodi.

[@aditya-u7s9v]

Beautiful solution sir!

[@MudithaMath]

Thanks Aditya!

[@anudeep0129]

noice, another approach wud be the show that M lies on the radical axis of circles (EA) and (AF) with centres being D and G respectively. motivation being:- noticing two nice circles and invoking the perp conditions of radical axis

[@MudithaMath]

@anudeep0129: Oh! I hadn't thought of that approach. But how do you show the M lies on the radical axis?

[@anudeep0129]

@@MudithaMath o lmao so i thought i had a proof but turns out i fakesolved, i'll try to continue with this approach and get back

[@anudeep0129]

@@MudithaMath so i asked one of my frnds abt this and he said he cudn't see any nice synthetic solution, but to trig bash and show that MD^2-DA^2=MG^2-GA^2 hence showing M lies on the radical axis.

[@viswanathan3992]

cool solution man!

[@MudithaMath]

Thanks :)