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It is proved that △ AED is equal to △ ACD if AE = AC is intercepted on ab
ν = ∠ AED = 2 ∠
∴∠B=∠BDE ∴BE=DE
∴AB=AE+BE=AC+CD

Given the triangle ABC, AB is greater than AC, try to explain that angle c is greater than angle B

Take a point D on AB to make ad = AC and connect to DC
Then: Triangle ADC is isosceles triangle
Angle ADC = angle ACD
But: angle ACB = angle ACD + angle BCD > angle ACD
In addition, the angle ADC is the outer angle of the triangle BCD
So: angle ADC = angle B + angle BCD > angle B
So: angle ACB > angle B

In triangle ABC, angle B = 2, angle a, BC = 2, ab = 4. Try to prove that triangle ABC is a right triangle

It is proved that: take point E on the extension line of AB, make be = BC = 2, connect CE, take the midpoint D of AB, connect CD ? be = BC = 2 ? e ? e ? ECB = 2 ∠ e ? e ? ECB = 2 ? e ? D is the midpoint of AB, ab = 4  BD = ad = AB / 2 = 2 ｡ BCD = BDC, ad = be

In the triangle ABC, the angle B is equal to 60 ° and ab = 2BC, which proves that the triangle ABC is a right triangle So can I draw a 30 ° right triangle myself and do it like it?

AB = 2bcab / sinc = BC / sina2bc / sinc = BC / sinasinc = 2sina

As shown in the figure, in the triangle ABC, ∠ B = 2 ∠ a, ab = 2BC, it is proved that the triangle ABC is a right triangle Figure: a It's three o'clock. C's fine 1 angle c is 90 degrees. I can see that

Two friends upstairs, how can we use Sina to solve this problem? A single Sina can only happen in RT △ and there is Sina = BC / ab. now we need to find △ why RT △ but you have already regarded it as RT △ what else can we ask for?
Let's see if I can explain it like this
Solution: take the midpoint o of BC as the center of the circle, 1 / 2Ab as the radius to draw a circle, and then draw a circle with point B as the center and 1 / 2Ab as the radius
∵ BC=1/2AB（AB=2BC）
The two circles must intersect at point C,
Ψ C = 90 (is the diameter of circle O and the circumference angle)
Ψ Δ ABC is RT △) (end)
(here's another proof.)
Take the midpoint o of BC as the center of the circle and 1 / 2Ab as the radius to draw a circle, and then draw a circle with point B as the center and 1 / 2Ab as the radius, i.e. BC = Bo = 1 / 2Ab -------- - 1)
Then co = Bo = 1 / 2Ab (radius of the same circle is equal) - 2
Comparison 1), 2)
∴BC=BO=CO
Δ CBO is equilateral
∴∠B=60 , ∠A=1/2∠B=30 , ∠C=90
Ψ Δ ABC is RT △) (end)

If the three sides of a triangle a, B, C satisfy the condition | A-12 | + B's square + C's square + 194 = 10B + 26c Please

|A-12| + B's square + C's square + 194 = 10B + 26c. →
|A-12 | + (square of b-10b + 25) + (square of c-26c + 169) = 0 →
|A-12 | + (B-5) squared + (C-13) squared = 0 →
a=12,b=5,c=13
The square of C = the square of a + the square of B
The ABC is a right triangle

As shown in the figure, in △ ABC, ∠ BAC = 2 ∠ B, ab = 2Ac, it is proved that △ ABC is a right triangle

It is proved that: as shown in the figure, the vertical bisector of line AB is D, and intersects with BC at point e. it is easy to prove that △ AED ≌ △ bed.  ad = 12ab = 12 × 2Ac = AC, ∠ B = ∠ EAD. ? BAC = 2 ∠ B, ∠ EAD + ∠ EAC = ∠ EAD

As shown in the figure, in △ ABC, ∠ BAC = 2 ∠ B, ab = 2Ac, it is proved that △ ABC is a right triangle

It is proved that, as shown in the figure, make the vertical bisector of line AB, and the perpendicular foot is D, and intersect with BC at point E, it is easy to prove △ AED ≌ △ bed
∴AD=1
2AB=1
2×2AC=AC，∠B=∠EAD．
∵∠BAC=2∠B，∠EAD+∠EAC=∠BAC，
∴∠EAC=∠EAD．
In △ AEC and △ AED, AE = AE, EAC = ead, AC = ad,
∴△AEC≌△AED．
∴∠C=∠EDA．
∵∠EDA=90°，
∴∠C=90°．
So △ ABC is a right triangle

In the triangle ABC, the angle BAC = 2, angle B, ab = 2Ac, AE bisect the angle BAC. Try to explain that the triangle ace is a right triangle

Angle B = angle BAE, deduce AE = be, take the midpoint of AB as h, connect eh, according to AE = be, the midpoint is h, so EH is perpendicular to AB, ah = 0.5ab = AC, angle BAE = angle CAE, so AEH of triangle is equal to ace of triangle, and it is not used later

In △ ABC, ad ⊥ BC is in D, ∠ B = ∠ DAC, then △ ABC is a right triangle. Please explain the reason

In ▷ abd, ▷ ADC
∠B=∠DAC
∠ADB=∠ADC
Ad = ad (common side)
▷ abd is all equal to ▷ ADC (AAS)
∴∠BAD=∠C
∵∠ADC=90°=∠DAC+∠C
∴∠DAC+∠DAB=90°
▷ ABC is the RT triangle