The first triangular problem, urgent! The first triangular problem

1. The algebraic formula b -c +2ac-a is equal to b -(c-a). It is known that a+b c introduces b c-a and, a, b, and c are all positive.

So the value of b greater than (c-a) b -c +2ac-a is positive.

2. You can refer to the corner edge formula, and the specific formula is given to the king. o( o alas.

3, a=180- eac=180-2 b=180-2 c ad bisects eac so dac= c so ad parallel bc

>0.Because any triangle is on both sides and greater than the third side.

2 Using the above formula, let c=2, a=7, by the title 20,b>5 so b=63 ad is parallel to bc, because cae= b+ c=2 b;

Since ad is an angular bisector, then ead=;

So AD is parallel to BC

As AE BC, right-angle AEC and right-angle BCD are co-C. Then EAC= DBC.

According to the three-in-one easy to know EAC = half of the BAC. Proven.

This question is wrong, to meet the conclusion, it is obvious that AD=DC is needed, but this condition does not seem to be obtainable, and it has nothing to do with the integration of three lines.

What are you asking for?? Issue? How do you ask for a title?

1. As shown in the figure, ea ab, bc ab, ab=ae=2bc, d is the dry noise point in ab, and the following four judgments are made: (1) de=ac; Yes, because ab=ae=2bc, d is ab midpoint, ad=1 2ab, abc congruence adeSo de=ac

2）de⊥ac；That's right, because a small = e, e + d=90, so f=90 so de ac

3）∠cab=30°；No, because bc=1 2 ab is not bc = 1 2 ac

4) Socks defeated eaf= ade is correct? Yes EAF + A Small = ADE + A Small = 90

Every judgment has a reason).

2. As shown in the figure, in RT ABC, ACB=90°, A B, M is the midpoint of the hypotenuse, the triangle ACM is folded along CM, and the point A falls at the point D. If cd happens to be perpendicular to ab, then a=?(Again, ask for the process and the answer).

a=∠d.Because ac=cd,cm=cm, acm= dcm,so acm congruent dcm,3, as shown in the figure, cd is the height on the hypotenuse of rt abc, bcd is folded along cd, b point falls exactly at the midpoint e e of ab, then a is equal to ?? (Process process, process is the most important, the answer is 30°).

That's right. Because EC is the midline on the hypotenuse of RT ABC=1 2AB,,EC=BC, BCE is congruent so A=30

4. As shown in the figure, it is known that in RT ABC, ACB=90°, CD AB, CE is the midline on the AB side, and BCD=3 DCAVerify: de=dc(Needless to say, everyone knows the result, the process).

Since CE is the midline on the AB side, CE=EB, B=C1, ESmall=2 B

and acd= b

So dce=3 dca- caubridgedca=2 dca=2 b

So dce= e small. So de=dc

de=1, so cd=2 so ac=2*2 root number 3, bc=ac2 root number 3=8 3

To the process, to the process).

2. As shown in the figure, in RT ABC, ACB=90°, A B, M is the midpoint of the hypotenuse, the triangle ACM is folded along CM, and the point A falls at the point D. If cd happens to be perpendicular to ab, then a=?(Again, ask for the process and the answer).

a=∠d.Because ac=cd,cm=cm, acm= dcm,so acm congruent dcm,3, as shown in the figure, cd is the height on the hypotenuse of rt abc, bcd is folded along cd, b point falls exactly at the midpoint e e of ab, then a is equal to ?? (Process process, process is the most important, the answer is 30°).

That's right. Because EC is the midline on the hypotenuse of RT ABC=1 2AB,,EC=BC, BCE is congruent so A=30

4. As shown in the figure, it is known that in RT ABC, ACB=90°, CD AB, CE is the midline on the AB side, and the slow bond scatters BCD=3 DCAVerify the perturbation: de=dc(Needless to say, everyone knows the result, the process).

Since CE is the midline on the AB side, CE=EB, B=C1, ESmall=2 B

and acd= b

So dce=3 dca- dca=2 dca=2 b

So dce= e small. So the bright code de=dc

de=1, so cd=2 so ac=2*2 root number 3, bc=ac2 root number 3=8 3

1.Be. Because the angular ECD is 45 degrees.

2.Be. Because both are right triangles.

3.Establish. Because it's the same as the second question.

4.is a fixed value. 180

EC bisects ACD, so ECD=45°, and ECB=90°, so CD bisects ECB

Solution: It can be obtained from the same area:

2s=ad*bc=be*ac

So 3*bc=4*ac

So ac:bc=3:4