Let I be incenter of triangle ABD os the square ABCD, E the intersection of DI with AB and F the contact of the incircle with AD. Prove that FI is the angle bisector of <CFI.
Proof:
Easy angle calculation gives that triangles CDI and AEI are isosceles, so EA/CD=AI/CI=AF/FD, i.e. AE/CD=AF/DF, relation proving that triangles AEF and DCF are similar, done.
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