sunmoonlight
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This reasoning is incorrect. You should point out that the quadrilateral so formed is a rhombus (all 4 sides equal, in this case = 1), whose diagonals bisect the angles of the rhombus. In general, diagonals of a parallelogram do not bisect their incident angles.Note that the lines from O to 1, from O to z and from z to z+1 form a parallelogram. The result follows immediately from the parellelogram property that the diagonals bisect the corner angles.
unfortunately this isn't true, it must be a rhombus not just a parallelogram.A rhombus is a parallelogram, just as a square is a rectangle. Showing it's a parallelogram is sufficient to establish the angle bisect property.
is this ext 2 math?View attachment 41531How to do this?
yeahis this ext 2 math?