For reference: If ABCD is a symmetric trapezoid, (AB = BC); M, N, L, F and P are midpoints of AB, BC, CD, AD and BD respectively. Calculate x (answer $36^\circ$)
My progress:
Symmetric trapezoids are those that have sides congruent 2 to two
IF AB = BC then AD=DC
MNFL is rectangle
$ K = ML\cap NF \\Draw~ MN \parallel FL\\\triangle FKL \cong\triangle MKN $
but I think it needs to demonstrate that the distance from HI = PG…
Best Answer
In triangle BNI, H is midpoint of BI. PL connects the midpoints of DB and DC so it is parallel with BN. Also HN||GL and HN=GL, in this way $\triangle BHN=\triangle BGL$ therefore $GP=HB=HI$. Also trapezoid PLNI is isosceles and extension of sides NI and PL intersect at a point like R such that triangle RNL is isosceles and RN is perpendicular bisector of NL so it is the axis of symmetry of trapezoid. Hence NI and PL are mirrors about RN or are symmetric of each other which results in $x=36^o$