Eight on the congruent triangle math problem 20 needs to be proved twice congruent

As far as I can judge, your situation may look like this:

There are math problems that need to prove congruent triangles twice, and it may be like this. For the first time, one condition or several conditions are demonstrated. The second time a condition or several pieces are proved, and the conditions of these two proofs are added up to meet the requirements of the topic, that is, to prove the congruent triangle of two congruences.

Make EP AB

EPA = EPB = 90° (definition of vertical).

d = c = 90° (known).

EPA= D, EPB= C (Equal Substitution) AE bisects DAE, BE bisects ABC

1 = 2, 3 = 4 (definition of angular bisector).

In AED vs. AEP.

EPA= D (verified).

1 = 2 (verified).

ae = ae (common edge).

aed≌△aep（aas）

In BEC and BEP.

epb = c (verified).

3 = 4 (verified).

ae = ae (common edge).

bec≌△bep（aas）

ad=ap (property of congruent triangles).

bc=bp (property of congruent triangles).

There is a diagram available: ab=ap+bp

ab=ad+bc

AED AEP, BEC BEP (Proven) DE=EP (property of congruent triangles).

ec=ep (property of congruent triangles).

DE=EC (Equivalent Substitution).

E is the CD midpoint (definition of the midpoint).

EF AD, so f= bad (equal isotope angles) and AEF= CAD (equal internal wrong angles).

AD bisects bac over bc to d, so cad = bad

In summary, f= bad= cad= aef

AEF is an isosceles triangle.

Proof: AD deuces the BAC

bad=∠cad

ef∥adefa=∠bad，∠cad=∠aef∴ ∠efa=∠aef

AEF is an isosceles triangle.

As shown in the figure, in the triangle ABC, the known angle ABC=45 degrees, CD is perpendicular to the point D, be bisects the angle ABC, and Be perpendicular AC is at the point E, intersects with CD at the point F, H is the midpoint on the edge of BC, and connects DH with the intersection point G

1.Verification: bf=ac

2.Verification: CE = 1/2 BF

3.Guess the quantitative relationship between CE and BG and prove your conclusions.

1. If the heights of the two sides and one of the two triangles are equal, then the angle (a) of the third side of the two triangles is opposite. Equal )a. Equal b. Not equal c. Mutual surplus d. Complementary or equal.

Two. In ABC and A'b'c', ab=a'b',ac=a'c'to prove ABC A'b'c'There are four ways of thinking to prove it:

1、bc=b'c' 2、∠a=∠a' 3、∠b=∠b' 4、∠c=∠c'

a、1234 b、234 c、12 d、34

c, 1 2, reason, triangle congruence can be equal on three sides, or it can be an angle on both sides. So it's 1 2

First, choose D, equal or complementary, and the other triangle may be an obtuse triangle.

Second, choose C, SSS, SAS

The added condition is ab ef

So bc=fd, ab=ef

Therefore, ABC is fully equal to EFD

ac=ed (sss)

b=∠f(sas)

a=∠e=90°(hl)

Whatever the conditions are, that's what it ends up looking like.