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Question 3 (5 pts) Soit le carré ABCD dont le côté mesure 2a. Soit F, le point milieu du côté AB et E, un point sur le côté AD tel que la mesure du segment AE est X. Exprimez x, en termes de a, pour que le triangle EFC soit rectangle en F.

Question 33 (55 pts)\newlineSoit le carré ABCD \mathrm{ABCD} dont le côté mesure 22a. Soit F \mathrm{F} , le point milieu du côté AB \mathrm{AB} et E \mathrm{E} , un point sur le côté AD \mathrm{AD} tel que la mesure du segment AE \mathrm{AE} est X \mathrm{X} .\newlineExprimez x x , en termes de a, pour que le triangle EFC soit rectangle en F F .

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Q. Question 33 (55 pts)\newlineSoit le carré ABCD \mathrm{ABCD} dont le côté mesure 22a. Soit F \mathrm{F} , le point milieu du côté AB \mathrm{AB} et E \mathrm{E} , un point sur le côté AD \mathrm{AD} tel que la mesure du segment AE \mathrm{AE} est X \mathrm{X} .\newlineExprimez x x , en termes de a, pour que le triangle EFC soit rectangle en F F .
  1. Triangle EFCEFC Formation: We know that triangle EFCEFC will be a right triangle at FF if EFEF is perpendicular to FCFC. Since FCFC is a horizontal line segment, EFEF must be a vertical line segment. This means that EE must be directly above FF on the side ADAD.
  2. Length of Line Segments: The length of ABAB is 2a2a, and FF is the midpoint of ABAB, so the length of AFAF is aa. Since EFEF is vertical and FCFC is horizontal, triangle EFCEFC is a right triangle with legs EFEF and FCFC.
  3. Condition for Right Triangle: For triangle EFCEFC to be a right triangle at FF, the length of AEAE must be equal to the length of AFAF, which is aa. So, xx, the length of AEAE, must be equal to aa.

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