In the isosceles right triangle ABC, AC = BC, ∠ C = 90 degrees, ad bisects ∠ BAC, De is perpendicular to the point e AB = 15 cm, and the circumference of the triangle DBE is calculated In the isosceles right triangle ABC, the angle c is 90 degrees, AC = BC, ad bisection angle BAC, de vertical AB, perpendicular foot e, and ab = 15 cm?

Because in the triangle ABC, the angle c = 90 degrees, AC = BC, and ab = 15, AC ^ 2 + BC ^ 2 = AB ^ 2, ab = BC = 15, root 2/
2. Because ad bisects the angle BAC, and DC is vertical to AC, De is perpendicular to AB, so DC = De, AE = AC, so the circumference of triangle DEB is de + DB + be = BC + (AB-AC) = 15

As shown in Fig. 2, in △ ABC, ∠ C = 90 °, AC = BC, ad bisect ∠ cab intersecting BC at point D, de ⊥ AB at point E. if AB = 8cm, then the circumference of △ DBE (done in the 51 issue of Zhejiang Education Press)

Where is figure 2?

As shown in the figure, in △ ABC, ∠ C = 90 ° AC = BC, ad is the bisector of ∠ BAC, de ⊥ AB, and the perpendicular foot is e. if AB = 10 cm, then the circumference of △ DBE is equal to () A. 10cm B. 8cm C. 12cm D. 9cm

∵ ad bisection ∠ cab, ∵ C = 90 °, de ⊥ AB,
∴CD=DE，
According to Pythagorean theorem, AC is obtained=
AD2−CD2，AE=
AD2−DE2，
∴AE=AC=BC，
∴DE+BD=CD+BE=BC，
∵AC=BC，
∴BD+DE=AC=AE，
The circumference of △ BDE is BD + de + be
=AE+BE
=AB
=10．
Therefore, a

As shown in the figure, in Rt △ ABC, AD ⊥ BC, Anton ABC=2 and Anton C are known. Try to explain the reason of AB+BD=CD

∵ RT △ ABC, ∵ B = 2 ∠ C,
∴∠B=60°，∠C=30°．
∴BC=2AB．
∵AD⊥BC，
∴∠BAD=30°．
∴AB=2BD．
∴BC=4BD
∴CD=3BD．
∴AB+BD=CD．

As shown in the graph, in RT △ ABC, ∠ C = 90 °, ad bisection ∠ BAC, CD = 1.5, BD = 2.5, find the length of AC

Do de by D, cross AB vertically, and cross AB at E
It is easy to prove that triangle ACD and AED are congruent, so de = CD = 1.5
Be = 2 can be obtained in right triangle DEB
Let AC be long x and have the equation x * x + 4 * 4 = (x + 2) (x + 2)
X=3

It is known that in △ ABC, ab = AC, BC = BD, ad = de = EB, then the degree of ∠ A is () A. 30° B. 36° C. 45° D. 50°

In this paper, we have the idea that ᙽ EBD = x

As shown in the figure, △ ABC is an isosceles right triangle, ab = AC = ad, ∠ CAD = 30 °. Calculate the degree of ∠ DCB ∠ DBC

∵AC=AD,∠CAD=30°
In the isosceles triangle, ∠ ACD = ∠ ADC = 75 °,
∵ △ ABC is a right triangle,
∴∠ACB=∠ABC=45°
∴∠BCD=∠ACD-∠ACB=75°-45°=30°
∵AB=AD,
∴∠DCB=∠BAC-∠CAD=90°-30°=60°
The △ abd is an equilateral triangle
∴∠ABD=60°
∴∠CBD=∠ABD-∠ABC=60°-45°=15°

As shown in the figure, in RT △ ABC, ∠ C = 90 °, AC = BC, be bisection ∠ ABC intersects AC with E, and a is used as the extension line of ad ⊥ be, and intersects point D. It is proved that ad = 1 2BE．

Proof: extend AD and BC to f point, as shown in the figure,
∵ BD ⊥ AD and BD bisection ∠ ABC,
∴AD=FD，
∵∠FAC+∠AED=90°，∠CBE+∠CEB=90°，
∴∠FAC=∠CBE，
And ? FCA = ∠ ECB = 90 ° AC = BC,
∴△AFC≌△BEC，
∴AF=BE，
∴AD=1
2BE．

As shown in the figure RT triangle ABC, the angle a = 45 degrees, the angle c = 90 degrees, and the point D on the line segment BC, BDC = 60 degrees, ad = 1, then BD =? It's point D on line AC

Where is your D point? I can't copy the title

As shown in the figure, we know that in RT △ ABC, ∠ ACB = 90 °, AC = 6cm, BC = 8cm, draw an arc with C as the center of the circle and Ca as the radius, and find the length of AD by crossing the oblique edge AB at point D The key to the process

∠C=90,AC=6,BC=8---->AB=10
Take the ad midpoint e --- > CE ⊥ AB --- > CE = AC * BC / AB = 4.8
--->AE = 3.6 --- > ad = 2ae = 7.2cm answer: Saker 2007-01-10 11:39 make the vertical line of AB cross AB at point E, connect CD, then △ ACD is isosceles triangle
AE=ED
In the right triangle ABC:
CE * AB / 2 = AC * BC / 2 (area of triangle ABC)
CE=AC*BC/AB=6*8/10=4.8
In a right triangle AEC:
AE^2=AC^2-CE^2=6^2-4.8^2
AE=3.6
So ad = 2ae = 7.2cm