Question

Prove that line segments joining the midpoints of consecutive sides of a rhombus form a triangle

Answer 1

draw rhombus ABCD with point A at the top and point C at the bottom

point D is at the left and point B is at the right

draw the segment joining the midpoints of AB and AD

label the midpoint of AB point F on the right

label the midpoint of AD point E on the left

points E and F are the midpoints of AD and AB

draw diagonal AC which intersects segment EF at point G

you should now see 2 triangles at the top of the diagram

these are triangle AFG and triangle AEG

these 2 triangles are congruent by ASA (angle-side-angle)

AF is congruent to AE because AB is congruent to AD and AF is half of AB and AE is half of AD

<AFG is congruent to <AEG because triangle AFE is isosceles and these are the base angles

again AF is congruent to AE

<GAF is congruent to <GAE because the diagonal of the rhombus bisects <EAF  (true for all parallelograms)

triangle AFG is congruent to triangle AEG by ASA again

<AGF is congruent to <AGE because corresponding angles of congruent triangles are congruent

but <EGF is a straight (180º) angle made up of <AGF+<AGE

if <AGF is congruent to <AGE and they sum to 180º, then <AGF and <AGE must both be 90º

if <AGF and <AGE are each 90º, then segment AC (the diagonal) and segment EF (the bisector of AB and AD) must be perpendicular because perpendicular lines intersect and form 90º angles

hope you understand all this; shorten this anyway you like

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