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As shown in the figure, in △ abd and △ ace, ab = ad, AC = AE, ∠ bad = ∠ CAE, connecting BC and de intersecting at point F, BC and ad intersecting at point g. verification: BC = De
Proof: ∵ bad = ∵ CAE,
∴∠BAD+∠DAC=∠CAE+∠DAC，
In other words, ∠ BAC = ∠ DAE
In △ cab and △ EAD
AB＝AD
∠BAC＝∠DAE
AC＝AE ，
∴△CAB≌△EAD（SAS），
∴BC=DE．
Given △ ABC, we make △ abd and △ ace with AB and AC as edges respectively, and ad = AB, AC = AE, ∠ DAB = ∠ CAE, connect DC and be, G and F are DC and B respectively (2) as shown in Figure 3, if ∠ DAB = x, try to explore the quantitative relationship between ∠ AFG and X, and give proof
Certificate: ∵ DAB = ∠ CAE
∴∠DAC = ∠BAE
Ad = AB, AC = AE
∴△DAC ≌△BAE
∴DC = BE,∠ADC = ∠ABE
G and F are the midpoints, DG = BF,
∴△DAG ≌△BAF
∴∠DAG = ∠BAF
∴∠GAF = ∠DAB =X
∴ ∠AFG=（180-X）/2
Given △ ABC, we make △ abd and △ ace with AB and AC as the edges, and ad = AB, ∠ DAB = ∠ CAE, connect DC and be, G, f are the midpoint of DC and be respectively
Certificate: ∵ DAB = ∠ CAE
∴∠DAC = ∠BAE
Ad = AB, AC = AE
∴△DAC ≌△BAE
∴DC = BE,∠ADC = ∠ABE
G and F are the midpoints, DG = BF,
∴△DAG ≌△BAF
∴∠DAG = ∠BAF
∴∠GAF = ∠DAB =X
∴ ∠AFG=（180-X）/2
Given the triangle ABC, ab = AC, ad is perpendicular to D, the circumference of triangle ABC, abd is 20cm, 16cm
Because AB = AC, so the circumference of ABC is ab + AC + BC, abd perimeter is ab + BD + ad, and because 2bd = BC, let ad be x, then we can calculate: ab + AC + BC = AB + AB + 2bd = 2Ab + 2bd = 20cm ① AB + BD + x = 16 ② ② × 2 = 2Ab + 2bd + 2x = 16 × 2 = 32 (3) - ① 2x = 32-20 = 12, x = 12 △ 2 = 6, that is ad = 6
As shown in the figure, in △ ABC, CD is the median line, BC-AC = 5cm, the circumference of △ DBC is 25cm, calculate the circumference of △ ADC
∵ CD is the center line,
∴AD=BD，
The circumference of △ DBC - △ ADC = (BC + BD + CD) - (AC + AD + CD) = BC-AC,
∵ BC-AC = 5cm, △ DBC perimeter is 25cm,
The circumference of 25 - △ ADC = 5,
It is found that the circumference of △ ADC is 20 cm
In △ ABC, ad is the center line on the edge of BC, the perimeter of △ ADC is 5cm more than that of △ abd, and the sum of AB and AC is 11cm, then the length of AC is______ .
As shown in the figure, ∵ ad is the center line on the BC side,
∴BD=CD，
∵ △ ADC perimeter - △ abd perimeter = ac-ab = 5,
And ∵ AB + AC = 11,
∴AC=5+11
2=8cm．
So the answer is: 8cm
In △ ABC, ad is the center line on the edge of BC, the perimeter of △ ADC is 5cm more than that of △ abd, and the sum of AB and AC is 11cm, then the length of AC is______ .
As shown in the figure, ∵ ad is the center line on the BC side,
∴BD=CD，
∵ △ ADC perimeter - △ abd perimeter = ac-ab = 5,
And ∵ AB + AC = 11,
∴AC=5+11
2=8cm．
So the answer is: 8cm
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As shown in the figure, ∵ ad is the center line on the BC side,
∴BD=CD，
∵ △ ADC perimeter - △ abd perimeter = ac-ab = 5,
And ∵ AB + AC = 11,
∴AC=5+11
2=8cm．
So the answer is: 8cm
As shown in the figure, in triangle ABC, D is the midpoint of BC. The perimeter of triangle ADC is 5cm more than that of ABD. The sum of AB and AC is 11cm. Find the length of AC
In the title: "D is the midpoint of BC, the perimeter of triangle ADC is 5cm more than that of abd", that is, AC is 5cm longer than AB, and because AC + ad = 11, AC = 8cm
If AB = 5cm, BC = 6cm, and midline ad = 4cm on BC side, then ∠ ADC=____ .
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