It is known that: as shown in the figure, one side De of rectangular defg is on the side BC of △ ABC, the vertices g and F are on the sides AB and AC respectively, ah is the height of the side BC, ah and GF intersect at the point K, BC = 12, ah = 6, EF: GF = 1:2 are known, and the perimeter of rectangular defg is calculated

It is known that: as shown in the figure, one side De of rectangular defg is on the side BC of △ ABC, the vertices g and F are on the sides AB and AC respectively, ah is the height of the side BC, ah and GF intersect at the point K, BC = 12, ah = 6, EF: GF = 1:2 are known, and the perimeter of rectangular defg is calculated

Let EF = x, then GF = 2x. ∵ GF ∥ BC, ah ⊥ BC, ≁ AK ⊥ GF. ∵ GF ∥ BC, ∥ AGF ∥ ABC, ∥ Akah = gfbc. ∵ ah = 6, BC = 12, ≁ 6 − X6 = 2x12. The solution is x = 3. The perimeter of rectangular defg is 18

In △ ABC of area 24, the edge De of rectangular defg moves on AB, and points F and G move on BC and AC respectively
(1) If AB = 8, de = 2ef, find the length of GF; (2) if ∠ ACB = 90 °, as shown in Figure 2, the line segments DM and en are the angular bisectors of △ ADG and △ bef respectively, and prove: Mg = NF; (3) please write the maximum area of rectangular defg directly

(1) Let EF = x, then GF = de = 2x, ∵ GF ∥ AB, ∥ CGF ∥ cab, ∥ gfab = h-efh, i.e. 2x8 = 6-x6, the solution is: x = 2.4, ∥ GF = 4.8; (2) GP ∥ BC over g, DP ∥ en over D, GP and DP intersect at P

As shown in the figure, in the angle ABC, the angle ABC is 90 degrees, and the square is made with the edge outward, and the area is recorded as S1, S2, S3 respectively. If S2 = 4, S3 = 6, then S1=
BC,AB,AC

Because triangle ABC is a right triangle with angle ABC = 90 degrees, according to Pythagorean theorem, there are ab ^ 2 + BC ^ 2 = AC ^ 2, and square S1 = BC ^ 2, square S2 = AB ^ 2, square S3 = AC ^ 2, so S1 + S2 = S3, S1 = s3-s2 = 6-4 = 2

Be ⊥ ad, DF ⊥ BC, and AO = Co ∠ ABC = CDA

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