Guangdong Learning Guidance Exercise Junior High School Grade 8 People's Education Edition Mathematics 14 Page 1 1 Question

1) Since ∠AEF is a right angle, ∠BAE and ∠FEC are complementary angles of ∠AEB, which proves that;

(2) According to the properties of a square, it is easy to prove that AG=EC, ∠ Age = ∠ ECF =135; Adding the equilateral angles obtained by (1), two triangles can be judged to be congruent by ASA.

(3) In Rt△ABE, AE2 can be easily obtained according to Pythagorean theorem; According to the congruent triangles of (2), AE=EF, that is, △AEF is isosceles Rt△, so its area is half of AE2, and the solution is obtained.

(1) Proof: ∫∠AEF = 90

∴∠fec+∠aeb=90； ( 1)

In Rt△ABE ∠ AEB+∠ BAE = 90,

∴∠bae=∠fec； (3 points)

(2) It is proved that ∵G and E are the midpoint of AB side and BC side of square ABCD, respectively.

∴AG=GB=BE=EC, and ∠ age =180-45 =135;

Cf is the bisector of ∠DCH,

∠ECF = 90+45 = 135； (4 points)

In △AGE and △ECF, Ag = EC ∠ age = ∠ ECF =135 o ∠ gae = ∠ FEC? ;

∴△age≌△ecf； (6 points)

(3) Solution: From △ age △ ECF, AE = ef is obtained.

∫∠AEF = 90 degrees,