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It is known that: as shown in the figure, ∠ ACB = 90 ° in △ ABC, CD is high, CE bisects ∠ BCD, and ∠ ACD: ∠ BCD = 1:2, then CE is the center line on AB side, right? Give reasons
CE is the center line on the edge of ab
Reason: ∵ ACB = 90 °, ACD: ∠ BCD = 1:2,
∴∠ACD=30°，∠BCD=60°，
∵ CE bisection ∵ BCD,
∴∠DCE=∠BCE=30°，
∵CD⊥AB，∠ACD=30°，∠BCD=60°，
∴∠A=60°，∠B=30°，
∴∠A=∠ACD+∠DCE=∠ACE，∠B=∠BCE，
∴AE=EC，BE=EC，
∴AE=BE，
Therefore, CE is the center line of AB side
As shown in the figure, in △ ABC, ∠ B = 115 °, the vertical bisector De of AC side intersects with ab edge at point D, and ∠ ACD: ∠ BCD = 5:3, then the degree of ∠ ACB is______ Degree
∵ de bisects AC vertically,
∴CD=AD，
∴∠A=∠DCA．
And ∵ ACD: ∵ BCD = 5:3,
∴∠ACD：∠ACB=5：8．
And ∵ B = 115 °,
∴∠A+∠ACB=65°，
∴∠ACB=65×8
13=40°．
As shown in the figure, it is known that AD and AE are the midline and high line of △ ABC respectively, and ab = 5cm, AC = 3cm. Find the circumference of (1) △ ABC and △ ACD
The circumference of △ abd = AB + AD + BD
The circumference of △ ACD = AC + AD + CD
CD = BD, so the perimeter difference is the difference between AB and AC, which is 2
In △ ABC, it is known that ad AE is the middle line line line on △ ABC edge BC, ab = 7cm AC = 6cm. The difference between the circumference of △ abd and △ ACD is The area relationship between △ abd and △ ACD is as follows
The circumference difference is 1
Equal area
As shown in the figure, ad is the center line of △ ABC. Given that the circumference of △ abd is 25cm, and ab is 6cm longer than AC, the circumference of △ ACD is______ cm．
∵ ad is the center line on the side of BC,
∴BD=CD，
The difference between the circumference of △ abd and △ ACD = (AB + BD + AD) - (AC + AD + CD) = ab-ac,
The circumference of abd is 25cm and ab is 6cm longer than AC,
The circumference of ACD is 25-6 = 19cm
So the answer is 19
As shown in the figure, ad is the center line of triangle ABC. Given that the circumference of triangle abd is 6cm smaller than that of triangle ACD, what is the difference between AC and ab
AC+AD+CD=AB+AD+BD+6
AC-AB=AD+BD+6-(AD+CD)=BD-CD+6
And ad is the center line of the triangle ABC, so BD = CD
So ac-ab = 6
Given that ad is the median line of the triangle ABC and ab = 5cm, then the relationship between the circumference of the triangle abd and the triangle ACD is__ The area relation between triangle abd and triangle ACD is__ .
, equal,
As shown in the figure, in the triangle ABC, ad is the middle line, and the height de and DF of the triangle abd and triangle ACD are calculated respectively through the point D. if AB = 4cm, AC = 3cm, de + DF = 3.5cm, the length of DF is calculated
∵ D is the midpoint of BC
The areas of ⊿ abd and ⊿ ACD are equal
∴ （1/2)AB×DE = (1/2)AC×DF
Substituting the known conditions, we can get:
4DE=3DF
And de + DF = 3.5
By solving this system of quadratic equations, we get: DF = 2
As shown in the figure, ad is the angular bisector of triangle ABC, de and DF are the heights of AB side in triangle abd and AC side in triangle ACD respectively Verification of ad vertical bisection ef
(known) ∵ bad = (known) ∵ bad = ≓ CAD (definition of angular bisectoline) ∵ de
As shown in the figure, in △ abd and △ ACD, it is known that ab = AC, ∠ B = ∠ C. It is proved that ad is the bisector of ∠ BAC
Proof: connect BC,
∵AB=AC，
∴∠ABC=∠ACB．
∵∠ABD=∠ACD，
∴∠DBC=∠DCB．
∴BD=CD．
In △ ADB and △ ADC,
BD＝CD
AB＝AC
AD＝AD ，
∴△ADB≌△ADC（SSS），
∴∠BAD=∠CAD，
That is, ad is the bisector of BAC
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