As shown in the figure, it is known that the triangle ABC is an equilateral triangle. D and E are on sides BC and AC respectively, and CD = CE. Connect de and extend to point F, so that EF = AE, connect AF, be and Cf. if AB = 6, BD = 2dc, find the area of abef of quadrilateral~

As shown in the figure, it is known that the triangle ABC is an equilateral triangle. D and E are on sides BC and AC respectively, and CD = CE. Connect de and extend to point F, so that EF = AE, connect AF, be and Cf. if AB = 6, BD = 2dc, find the area of abef of quadrilateral~

Through a, Ag is perpendicular to BC, EH is perpendicular to BC, Ag = 3, root 3, eh = root 3, parallelogram abdf = 4 * root 3 = 12 root 3, triangle BDE = 0.5 * 4 * root 3 = 2 root sign 3, and the two formulas are subtracted to obtain the area of 10 root sign 3

As shown in the figure, in △ ABC, O is the intersection of high AD and be As shown in the figure, in △ ABC, O is the intersection point of high AD and be. Observe the graph and try to guess what kind of quantitative relationship exists between ∠ C and ∠ doe, and prove your conclusion

The sum of the inner angles of the quadrilateral CDOE is 360 degrees
∠C+∠DOE+∠CDO+∠CEO=360°
∠C+∠DOE+ 90°+ 90°=360°
∠C+∠DOE=180°

As shown in the figure, ad, be, CF are the three midlines of △ ABC, then AB = 2______ ，BD=______ ，AE=1 2______ .

∵ CF is the center line on the edge of ab,
∴AB=2AF=2BF；
∵ ad is the center line on the side of BC,
∴BD=CD，
∵ BE is the center line on the AC side,
∴AE=1
2AC，
So the answer is: AF; CD; AC

In the triangle ABC, if ad is the bisector of angle a, BD: DC = 2:3, AC = 6, then the length of the median line parallel to side AB is In the triangle ABC, if ad is the bisector of angle a, BD: DC = 2:3, AC = 6, then the length of median line parallel to ab side is () A.4 B.2 C.6 D.3

B. In the exclusion method, we can see the answer at a glance. Since BD: DC = 2:3, do db 'to AC to B', and ensure that the triangle abd and triangle AB'd are congruent, then we can see that ab '= AB, that is, ab = ab'

As shown in the figure, in △ ABC, ab = AC, D is a point outside △ ABC, and ∠ abd = 60 ° and ∠ ACD = 60 ° Confirmation: BD + DC = ab

It is proved that: extend BD to f so that BF = Ba, connect AF, CF, ? abd = 60 degrees,  ABF is an equilateral triangle,  AF = AB = AC = BF, ∠ AFB = 60 ° and ∵ ACF = ∠ AFC, and ? ACD = 60 °, ∵ AFB = ∠ ACD = 60 °, DFC = ∠ DCF,  DC = BD + DF = BF = AB, namely BD + D

As shown in the figure, in RT △ ABC, ∠ C = 90 ° D is a point on AC, and Da = DB = 5, and the area of △ DAB is 10, then the length of DC is () A. 4 B. 3 C. 5 D. 4.5

∵ in RT △ ABC, ∵ C = 90 °,
⊥ AC, that is, BC is the height of ⊥ DAB,
The area of DAB is 10, Da = 5,
∴1
2DA•BC=10，
∴BC=4，
∴CD=
DB2−BC2=
25−16=3．
Therefore, B

As shown in the figure, in RT △ ABC, ∠ C = 90 ° D is a point on AC, and Da = DB = 5, and the area of △ DAB is 10, then the length of DC is () A. 4 B. 3 C. 5 D. 4.5

∵ in RT △ ABC, ∵ C = 90 °,
⊥ AC, that is, BC is the height of ⊥ DAB,
The area of DAB is 10, Da = 5,
∴1
2DA•BC=10，
∴BC=4，
∴CD=
DB2−BC2=
25−16=3．
Therefore, B

In the triangle ABC, ab = AC, point D on BC, and BD = ad, DC = AC, find the degree of angle B, sorry, no figure

DC=AC,∠CAD=∠ADC
And ∠ ADC = ∠ DAB + ∠ B
And BD = ad, ∠ DAB = ∠ B
And ∠ C = ∠ B
That is ∠ ADC = ∠ CAD = 2 ∠ C
∴∠C+∠CAD+∠ADC=180°=5∠C
∴∠B=∠C=180/5=36°

If the area of △ ABC is 1 square centimeter, DC = 2bd, AE = 3Ed, then the area of △ ace is______ Square centimeter

∵S△ABC=1，DC=2BD，
∴S△ACD=2
3，
And ∵ AE = 3Ed,
∴S△ACE=3
4S△ACD=3
4× 2
3=1
2cm2．
So the answer is: 1
2．

If the area of △ ABC is 1 square centimeter, DC = 2bd, AE = 3Ed, then the area of △ ace is______ Square centimeter

∵S△ABC=1，DC=2BD，
∴S△ACD=2
3，
And ∵ AE = 3Ed,
∴S△ACE=3
4S△ACD=3
4× 2
3=1
2cm2．
So the answer is: 1
2．