It is known that in the parallelogram ABCD, points E and F are on AB and BC respectively. If △ ade, △ bef and △ CDF are 5, 3 and 4 respectively, what is the area of △ def

It is known that in the parallelogram ABCD, points E and F are on AB and BC respectively. If △ ade, △ bef and △ CDF are 5, 3 and 4 respectively, what is the area of △ def

First of all, if the quadrilateral area is fixed, the area of triangle ade is only related to AE / AB, and the same is true for other triangles, so I can set ABCD as a rectangle and let ad = 5, then AE = 2, be = x BF = 6 / X CF = 4 / (x + 2) 6 / x + 4 / (x + 2) = 5, so x = 2, then the area of triangle DEF is 8

In the parallelogram ABCD, the bisector of the angle bad intersects the straight line BC at point E and the straight line DC at F
In &; ABCD, the bisector of ∠ bad intersects line BC at point E and intersects line DC at point F
Q: if ∠ ABC = 120 °, FG / / CE, FG = CE, connect dB and DG respectively (as shown in the figure), and calculate the degree of ∠ BDG

Because ab ∥ DC ∥ BAF = ∠ DFA and ∥ DAF = ∠ BAF ∥ DAF = ∠ DFA ∥ Da = DF, because ad = BC ∥ DF = BC connected to BG, GF.AB Because of ∠ BAE = ∠ DAF ≠ CFE = ∠ CEF ∈ CE = CF and CE = FG} CF = FG, because of ad ‖ BC} FG}}}}}} CF = FG}} CF} CF} CF} CF} CF} CF} CF} CF} CF} CF} CF} CF} CF} CF} CF} CF} CF} CF}

In the parallelogram ABCD, the bisector of ∠ bad intersects the straight line BC at point E, the extension of the intersecting straight line DC at point F, and takes EC and CF as the adjacent sides as the parallelogram

In the parallelogram ABCD, the bisector of ∠ bad intersects the straight line BC at point E and the intersecting straight line DC at point F. (1) it is proved in Fig. 1 that CE = CF;
(2) If ∠ ABC = 90 ° and G is the midpoint of EF (as shown in Figure 2), write the degree of ∠ BDG directly;
(3) If ∠ ABC = 120 ° FG ‖ CE, FG = CE, connect dB and DG respectively (as shown in Figure 3), and calculate the degree of ∠ BDG
⑴∵AD∥BC
∴∠DAE=∠AEB
∵AB∥CD
∴∠BAE=∠CFE
∵ AE bisection ∠ bad
∴∠BAE=∠DAE
∴∠CEF=∠AEB=∠CFE
∴CE=CF
⑵45°
Connect BG, CG
∵BE=AB=DC
EG=CG
∠BEG=135°=∠DCG
∴△BEG≌△DCG,BG=DG
∴∠BGE=∠DGC
∴∠BGD=∠EGC=90°
{△ BDG is an isosceles right triangle
∴∠BDG=45°
(3) connect BG and CG
It is easy to prove that CEGF is rhombic
ABC = 120 degree
∴EG=CG
In addition, be = AB = DC
∴△BEG≌△DCG
∴BG=DG,∠BGE=∠DGC
∴∠BGD=∠EGC=60°
Ψ△ bgd is an equilateral triangle
∴∠BDG=60°
Give me the best! Typing is very tired!