As shown in the figure: ab = ad, CB = CD. △ ABC and △ ADC are congruent? Why?

As shown in the figure: ab = ad, CB = CD. △ ABC and △ ADC are congruent? Why?

Δ ABC and △ ADC are congruent,
The reasons are as follows.
∵ AB = ad, CB = CD and AC is the common edge,
∴△ABC≌△ADC（SSS）．

As shown in the figure: ab = ad, CB = CD. △ ABC and △ ADC are congruent? Why?

Δ ABC and △ ADC are congruent,
The reasons are as follows.
∵ AB = ad, CB = CD and AC is the common edge,
∴△ABC≌△ADC（SSS）．

As shown in the figure: ab = ad, CB = CD. △ ABC and △ ADC are congruent? Why?

Δ ABC and △ ADC are congruent,
The reasons are as follows.
∵ AB = ad, CB = CD and AC is the common edge,
∴△ABC≌△ADC（SSS）．

As shown in the figure: ab = ad, CB = CD. △ ABC and △ ADC are congruent? Why?

Δ ABC and △ ADC are congruent,
The reasons are as follows.
∵ AB = ad, CB = CD and AC is the common edge,
∴△ABC≌△ADC（SSS）．

To prove that the angles of ABC, ABC, CB are angles of the triangle

Proof: extend CE to F, make ef = CE, connect FB
∵ CE is the center line of ᙽ ABC,
∴AE=EB,
And ∵ AEC = ∵ bef,
∴△AEC≌△BEF,（SAS）
∴∠A=∠EBF,AC=FB．
∵AB=AC,
∴∠ABC=∠ACB,
∴∠CBD=∠A+∠ACB=∠EBF+∠ABC=∠CBF；
ADC is the center line of △ CB,
∴AB=BD,
And ∵ AB = AC, AC = FB,
∴FB=BD,
CB = CB,
∴△CBF≌△CBD（SAS）,
∴CD=CF=CE+EF=2CE

As shown in the figure, CE and CB are the midlines of △ ABC and △ ADC respectively, and ab = AC. verification: CD = 2ce

It is proved that: extend CE to F, make ef = CE, connect FB. ? CE is the center line of △ ABC, ? AE = EB, and ? AEC = ≌≌≌≌≌≌≌≌≌≌≌≌△ bef, (SAS) ? a = ∠ EBF, AC = FB. ? AB = AC,  ABC = ∠ ACB, ? CB is the midline of △ ADC

As shown in the figure CE, CB are the midlines of △ ABC and △ ADC, ab = AC, ∠ ABC = ∠ ACB, and CD = 2ce

It is proved that: extend CE to F, make ef = CE, connect FB. ? CE is the center line of △ ABC, ? AE = EB, and ? AEC = ∠ bef, ? AEC ≌≌△ bef, (SAS) ? a = ∠ EBF, AC = FB. ? AB = AC, ? ABC = ∠ ACB, ? CBD = ∠ a + ∠ ACB = ∠ EBF + ∠ ABC = ∠ CB is the midline of ADC, ? CB is the midline of ADC

As shown in the figure, CE and CB are the midlines of △ ABC and △ ADC respectively, and ab = AC. verification: CD = 2ce

Proof: extend CE to F, make ef = CE, connect FB
∵ CE is the center line of ᙽ ABC,
∴AE=EB，
And ∵ AEC = ∵ bef,
∴△AEC≌△BEF，（SAS）
∴∠A=∠EBF，AC=FB．
∵AB=AC，
∴∠ABC=∠ACB，
∴∠CBD=∠A+∠ACB=∠EBF+∠ABC=∠CBF；
∵ CB is the center line of ᙽ ADC,
∴AB=BD，
And ∵ AB = AC, AC = FB,
∴FB=BD，
CB = CB,
∴△CBF≌△CBD（SAS），
∴CD=CF=CE+EF=2CE．

In the triangle ABC, D is the point on BC, and BD ratio DC is equal to 2:1, s triangle ACD is equal to 12, s triangle ABC is equal to how many?

If the height of the triangle is h, then
S△ABC= S△ABD + S△ACD
=（BD * h）/2 + (CD * h)/2
=(2CD * h)/2 + (CD * h)/2
=3 S△ACD
=3 x 12
=36

In the triangle ABC, a, B, C are the opposite sides of angle a, angle B and angle C respectively. If a, B, C form an arithmetic sequence, the angle B = 30 degrees, and the area of triangle ABC is 0.5, then B is Here's my calculation: 1/2ac*sinB=1/2 a+c=2b b^2=a^2+c^2-2ac*cosB It's a strange number

Two formulas are needed
1、S=0.5ac*sinB
2. The square of B = the square of a + the square of C - 2 * a * c * CoSb
Determinant
0.5=0.5ac*sin30
The square of B = the square of a + the square of C - 2 * a * c * cos30
Find B = (radical 2) - 1