If there is a point D in the triangle ABC that connects Da, DB and DC, there are () triangles that do not overlap each other

If there is a point D in the triangle ABC that connects Da, DB and DC, there are () triangles that do not overlap each other

According to the drawing, D in the middle of the triangle ABC divides it into three triangles, namely: abd, ADC, BDC; according to the requirements of the title, there are several non overlapping triangles, and we can see that there are only four: the last one is the big triangle line ABC. So the answer should be four

Known: as shown in the figure, in the triangle ABC, ab = AC, D is a point in the triangle ABC, DB = DC, prove ∠ abd = ∠ ACD

ABC is an isosceles triangle with equal base angles. DBC is also an isosceles triangle with equal base angles

As shown in the figure, D is a point on the bisector of angle a of the triangle ABC, connecting dB and DC, and angle abd = angle ACD. Try to judge the position relationship between AD and BC

AD⊥BC
reason:
∵ ad bisection ∵ bac
∴∠BAD=∠CAD
∵∠ABD=∠ACD,AD=AD
∴⊿ABD≌⊿ACD﹙AAS﹚
∴AB=AC
∵ ad bisection ∵ bac
⊥ BC

As shown in the figure, △ ABC, ab = 2Ac, ∠ 1 = ∠ 2, Da = dB. Can you explain DC ⊥ AC?

As shown in the figure, make de ⊥ AB at E,
∵DA=DB，DE⊥AB，
∴AE=EB=1
2AB，∠AED=90°．
∵AB=2AC，
∴AC=1
2AB．
∴AC=AE．
In △ ACD and △ AED,
∵AC=AE，∠2=∠1，AD=AD，
∴△ACD≌△AED（SAS）．
∴∠ACD=∠AED=90°．
∴DC⊥AC．

As shown in the figure, known AB = AC, DB = DC, try to explain ∠ abd = ∠ ACD

Proof: ab = AC,
∴∠ABC=∠ACB．
∵BD=CD．
∴∠DBC=∠DCB．
∴∠ABC-∠DBC=∠ACB-∠DCB．
That is ∠ abd = ∠ ACD

As shown in the figure: it is known that in RT △ ABC, ∠ ACB = 90 °, CD is the median line on edge AB, AC = 6, cos ∠ ACD = 2 3. Find the length of ab

∵ in RT △ ABC, ∵ ACB = 90 ° and CD is the center line on edge ab,
∴AD=CD；
∴∠A=∠ACD，
∵cos∠ACD=2
3，
∴cos∠A=2
Three
∵cos∠A=AC
AB，AC=6，
∴6
AB=2
3，
∴AB=9，
So the length of AB is 9

In RT △ ABC, ∠ C = 90 ° CD ⊥ AB at point D, AC = 3, BC = 4, then what is tan ∠ ACD equal to?

∵∠C=90°,
∴∠A+∠B=90°
∵ CD ⊥ AB at point d
∴∠ADC=90°
∠A+∠ACD=90°
∴∠B=∠ACD
∴tan∠ACD=tan∠B=AC/CB=3/4

In RT △ ABC, the angle c is 90 ° and CD is perpendicular to D. the ratio of BD to ad is 1:4. The sine value and cosine value of angle B are obtained

Let BD = a, then ad = 4A
∵CD^2=BD*AD
∴CD=2a
∵CB^2=CD^2+BD^2
∴CB=√5a
sinB=CD/CB=2√5/5
cosB=BD/CB=√5/5

As shown in the figure, we know that in the RT triangle ABC, the angle c = 90 degrees, ab = BC, ad is the bisector of angle A

Condition error: ab = BC error, should be AC = BC
D as de ⊥ ABAB at E,
∵∠DAC=∠DAE,DC⊥AC,DE⊥AE,
Ad is the public side,
∴△ADC≌△ADE（AAS）
∴CD=ED,AC=AE,
AC + CD = AE + ed,
AC = BC, B = 45,
De = EB,
∴AC+CD=AE+EB=AB.
The proof is over

In the RT triangle ABC, if C = 90 degrees, CD is perpendicular to D, ad = 6, BD = 2, then BC=

According to the law of right triangle, CD squared = BD times ad, so CD = double root sign 3 can be obtained
From Pythagorean theorem, BC = 4