Attached are the question and solution for part (a) and (b).
For part (c), how to prove that EF is parallel to AC?
Thank you.
Hi Yeoky,
Complete the following constructions to your drawing
Extend BH to AD @ J BG to DC @K
connect JK,JE,KF,JE, EK
There are congruencies leading to BG= HD BH= GD BHDG is a parallelogram
Also EJ =FK EF=JK EJKF is a parallelogram
The perpendicular distance from AC to EF and JK are equal
EF is parallel to AC.
Find the congruent parts
Its very simple just prove that figure a parallelogram by joining its all sides and diagonals.This will prove that the difference of two lines is equal at every point.
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