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As shown in the figure, in the triangle ABC, the midlines BD and CE intersect at the points o, e and F are the midpoint of OB and OC respectively, indicating that the quadrilateral defg is a parallelogram
e. B and B are parallel to each other;
f. G is the midpoint of OB and OC respectively, so FG is parallel to BC;
Therefore, ED is parallel to FG
e. F is the midpoint of AB and ob respectively, so EF is parallel to AO;
d. G is the midpoint of AC and OC respectively, so DG is parallel to AO;
Therefore, EF is parallel to dg
If the opposite sides of defg quadrilateral are parallel, the quadrilateral is parallelogram
It is known that: as shown in the figure, the two heights BD and CE of the acute angle △ ABC intersect at point O, and ob = OC Verification: OA bisection ∠ BAC
Proof: CE ⊥ AB, BD ⊥ AC,
∴∠AEC=∠ADB=∠BEC=∠CDB=90°．
∵OB=OC，
∴∠DBC=∠ECB．
In △ BCD and △ CBE,
∠BEC＝∠CDB
∠BCE＝∠DBC
BC＝CB ，
∴△BCD≌△CBE（AAS），
∴BD=CE．
∵OB=OC，
∴BD-OB=EC-OC
∴OD=OE．
In RT △ ODA and RT △ OEA,
AO＝AO
OD＝OE ，
∴Rt△ADO≌Rt△AEO（HL），
∴∠DAO=∠EAO，
Ψ OA bisection ∠ BAC
It is known that: as shown in the figure, the two heights BD and CE of the acute angle △ ABC intersect at point O, and ob = OC (1) It is proved that △ ABC is an isosceles triangle; (2) Judge whether the point O is on the bisector of ∠ BAC and explain the reason
（1） Proof: ∵ OB=OC,
∴∠OBC=∠OCB，
∵ acute angle ᙽ the two heights BD and CE of ABC intersect at point o,
∴∠BEC=∠CDB=90°，
∵∠BEC+∠BCE+∠ABC=∠CDB+∠DBC+∠ACB=180°，
∴180°-∠BEC-∠BCE=180°-∠CDB-∠CBD，
∴∠ABC=∠ACB，
∴AB=AC，
The △ ABC is an isosceles triangle;
(2) Point O is on the angular bisector of ∠ BAC
Reason: connect AO and extend BC to F,
In △ AOB and △ AOC,
AB=AC
OB=OC
OA=OA
∴△AOB≌△AOC（SSS）．
∴∠BAF=∠CAF，
The point O is on the angular bisector of ∠ BAC
As shown in the figure, in RT △ ABC, ∠ C = 90 ° and the lengths of AB, BC and Ca are C, a, B respectively
Let the radius of the inscribed circle be r
∵S△ABC=1
2ab=1
2（a+b+c）•r，
∴r=ab
a+b+c．
As shown in the figure, in RT △ ABC, ∠ C = 90 ° and the lengths of AB, BC and Ca are C, a, B respectively
Let the radius of the inscribed circle be r
∵S△ABC=1
2ab=1
2（a+b+c）•r，
∴r=ab
a+b+c．
In RT △ C = 90 ° the lengths of AB, BC and Ca are C, a and B respectively. Find the radius r of the inscribed circle of △ ABC
Let the radius of the inscribed circle of △ ABC be r, its area be s, and its half circumference s = (a + B + C) / 2
Then, r = s / s, s = (1 / 2) * ab
Therefore, r = (1 / 2) * AB / (a + B + C) / 2
A: r = AB / (a + B + C)
As shown in the figure, in RT △ ABC, the area of △ ABC is as follows: ad = 5cm on the slope BC and AE = 6cm on the central line on the hypotenuse BC______ cm2．
∵ the center line AE on the beveled edge BC is 6 cm,
∴BC=12cm，
∵ the line ad on the slope BC is 5 cm,
The area of △ ABC = 1
2×12×5=30cm2．
So the answer is: 30
As shown in the figure, in the triangle ABC, the angle ACB = 90 degrees, CD is the height on the AB side, ab = 13cm, BC = 12cm, AC = 5cm, calculate the length of CD
CD = 60 / 13, angle ACB = 90 degrees, indicating that it is a right triangle, area is 12 * 5 = 60, area = bottom * height, bottom AB = 13, so high CD = area 60 / 13, I'm your friend in DNF, just a brother. It seems that you can't read well,
As shown in the figure, in re △ ABC, the angle ACB = 90 ° and CD is the height of AB edge, ab = 13cm, BC = 12cm, AC = 5cm. Find the length of area CD of △ ABC
1. Angle ACB = 90 °, AC = 5, BC = 12
S△ABC＝1/2（AC*BC）＝30
2、CD＝60/13
As shown in the figure, ∠ C = 90 ° in △ ABC, CA = CB, ad bisect ∠ BAC, de ⊥ AB, if AB = 6cm, calculate the circumference of △ DBE It is suggested that EF = EB should be intercepted on AE and DF should be linked
Intercept EF = EB on AE and connect DF
De vertical BF, be = EF, DF = dB, DF vertical dB, de = EF
Angle ADB = angle ACD + angle CAD = angle FDB + angle ADF
Angle CAD = angle ADF
The angle DAB = angle ADF
DF = AF
DBE perimeter = be + DF + DF = be + EF + AF = AB = 6
RELATED INFORMATIONS
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