Given Triangle ABC, let D be mid point of BC and let E and F be two point on sides AB and AC respectively such that angle EDF is right angle (90°). Prove that BE + CF > EF. I was able to do it by using triangle inequality but the book I am using has not yet introduced it (neither similarity of triangle) so I want much Elementary proof (like some construction).
My methods outline goes as follows Extend ED to R such that DR=DE now join point C and R and we have triangle DCR congurent to triangle DBE. By cpct we have RC = BE Now join point F and R now we have again we get triangle FED congurent to triangle FRD thus FR = EF by cpct. Thus from triangle DCR we have FC + CR >FR Or FC + BE > EF
Let $G$ be a mid-point of $EF$.
Thus, by the triangle inequality we obtain: $$EF=2|\vec{DG}|=|\vec{BE}+\vec{CF}|\leq|\vec{BE}|+|\vec{CF}|=BE+CF.$$ The equality occurs, when $BE||CF,$ which is impossible.