As shown in the figure, in △ ABC, de ‖ BC, ad: DB = 3:2 (1) find the value of DEBC; (2) find the value of bced & nbsp; of s △ ADEs quadrilateral

As shown in the figure, in △ ABC, de ‖ BC, ad: DB = 3:2 (1) find the value of DEBC; (2) find the value of bced & nbsp; of s △ ADEs quadrilateral

(1) ∫ de ∥ BC ∫ ade ∫ ABC ∫ ad: ab = de: BC ∫ ad: DB = 3:2, ∫ ad: ab = 3:5 ∫ de: BC = 3:5; (2) ∫ ade ∫ ABC, ∫ s △ ade: s △ ABC = 9:25, ∫ s △ ADEs quadrilateral bced = 916

In the triangle ABC, the angle BAC = 45 ° ad is perpendicular to D, the triangle abd is folded along the line where AB is located, so that point d falls at point E; the triangle ACD is folded along the line where AC is located
(1) judge the shape of quadrilateral aemf and give proof. (2) if BD = 1, CD = 2, try to find the area of quadrilateral aemf

(1) The ∵ ad ⊥ BC, △ AEB is obtained by ∵ ADB folding, ∵ 1 = ⊥ 3, ∵ e = ⊥ ADB = 90 °, be = BD, AE = ad. and ∵ AFC is obtained by ∵ ADC folding, ∵ 2 = ⊥ 4, ∵ f = ⊥ ADC = 90 °, FC = CD, AF = ad. ∵ AE = AF. (2 points) and ∵ 1 + ⊥ 2 = 45 °, ∵ 3 + ≁ 4 = 45 °

The three sides of the triangle ABC are AC = 5, BC = 12, ab = 13. Fold the triangle ABC along ad so that point C falls at point e of ab. what is the length of CD?
I don't have any reward points

Because the angle c is 90 degrees and the triangle is folded along ad, the angle DEA is 90 degrees, so the angle DEB is 90 degrees, so the square of de + the square of be is equal to the square of DB. Because CD is equal to de, let BD be x, then De is 12-x, because AE is 5, so be is 8. According to Pythagorean theorem, X is 26 / 3, CD is 10 / 3

In the equilateral triangle ABC, BD = 1 AB / 3, CE = 1 AC / 3, it is proved that De is perpendicular to AC

Is d on the extension line of AB
In the triangle ade, ad = four-thirds AB, AE = two-thirds AC, ab = AC, that is, ad = 2ae, angle a = 60 degrees. According to the triangle theorem, De is perpendicular to AC

In the triangle ABC, if angle a + angle B = angle c and AC = 1 AB / 2, then angle B=

Let the opposite sides of angles a, B and C be a, B and C respectively
AC=1/2AB
b=1/2c

As shown in the figure, in △ ABC, CE is the angular bisector, eg ‖ BC, intersects AC side at F, intersects the bisector of the outer angle of ∠ ACB & nbsp; (∠ ACD) at g, explores the quantitative relationship between EF and FG, and proves your conclusion

EF = FG, reason: ∵ CE is angle bisector, ∵ 1 = ∵ 2, ∵ eg ‖ BC, ∵ 3 = ∵ 1, ∵ 3 = ∵ 2, ∵ EF = CF, ∵ CG bisector ∵ ACD, ∵ eg ‖ BC, ? eg ‖ BC, ∵ eg ‖ 4 = ∵ g, ∵ FC = FG, ? EF = FG

In the triangle ABC, a = ACB, CD is the bisector of ACB, CE ⊥ AB in E indicates that CDB = 3 DCB

It is proved that let ACB = 2x
∵∠A＝∠ACB,∠ACB＝2X
∴∠A＝2X
∵ CD bisection ∠ ACB
∴∠DCB＝∠ACD＝∠ACB/2＝X
∴∠DCB＝∠A+∠ACD＝2X+X＝3X
∴∠CDB＝3∠DCB

It is known that be and CD are the high lines of △ ABC respectively, and BD = CE

According to the meaning of the title, BD = CE, ∠ BDC = ∠ BEC = 90 ° and BC is a right angle side, we can see that CD = be, △ BDC and △ BEC have the same area and are similar triangles, ∠ DBC = ∠ ACB, so AB = AC and △ ABC are isosceles triangles

In RT △ ABC, point m is the midpoint of edge AB, if BM = 2, what is cm
2x square-2 root sign 2x-5 = 0, solve the equation
It is known that the quadratic equation KX square-x-1 = 0 with respect to X has two unequal real roots?
3) It is known that the abscissa of an intersection point of the positive scale function y = KX image and the inverse scale function y = x (K + 3) image is 2?

In a right triangle, the center line of the hypotenuse equals half of the hypotenuse

In the triangle ABC, M is a point on BC, am is the bisector of the angle BAC, if AB = 2, AC = 1, BM = 1.5, then the length of CM is?
Urgent! Thank you!
Better explain the process! ~

Making CD am through C
Because CD ‖ am, AB / ad = BM / DC
Because am bisects ∠ BAC, so ∠ BAM = ∠ Mac
From CD ∥ am, we know that ∥ BAM = ∥ D, ∥ MAC = ∥ ACD
Then, d = ACD, that is, ad = AC
So AB / AC = BM / cm
BM/MC=AB/AC
1.5/CM=2/1
CM=0.75