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In the RT triangle ABC, if the angle ACB = 90 degrees, BC = 3, AC = 4, the vertical bisector De of AB intersects the extension of BC at point E, then
In RT Δ ABC, if ∠ ACB = 90 ° BC = 3, AC = 4, the vertical bisector De of AB intersects the extension of BC at point E, then the length of CE is
According to Pythagorean theorem, ab = 5
∵∠BDE=∠ACB=90°
∠B=∠B
∴△ABC∽△EBD
∴BD/BC =BE/AB
∴2.5/3=BE/5
∴BE=25/6
∴CE=25/6-3=7/6
As shown in the figure, in the RT triangle ABC, the angle ACB = 90 degrees, make the vertical bisector of AC, intersect AB with D, intersect AC with E, and connect CD (1) Try to explain that CD is the middle line on the edge of AB, and try to judge the quantitative relationship between Cd and ab (2) If AC = 12 cm and CD = 6.5 cm, find the area of the triangle ABC We need it now
(1) Because the angle ACB is 90 degrees
So ∠ a + ∠ B = 90 °, 1 + ∠ 2 = 90 °
Because the vertical bisector of AC is de
So AE = CE, ad = CD
So ∠ 1 = ∠ a
So ∠ B = ∠ 2
So CD = BD
So ad = BD
So CD is the center line on the edge of AB, CD = 1 / 2Ab
(2) Because AC = 12 cm, CD = 6.5 cm
So AB = 13 cm
BC = radical (13? - 12?) = 5
So the area of the triangle ABC is 12 × 5 △ 2 = 30 square centimeters
As shown in the figure, in RT △ ABC, the vertical bisector De of ∠ ACB = 90 ° intersects with the extension line of BC at F. if ∠ f = 30 ° de = 1, the length of EF is () A. 3 B. 2 C. Three D. 1
Connect AF,
∵ the vertical bisector De of AB intersects with the extension line of BC at F,
∴AF=BF，
∵FD⊥AB，
∴∠AFD=∠BFD=30°，∠B=∠FAB=90°-30°=60°，
∵∠ACB=90°，
∴∠BAC=30°，∠FAC=60°-30°=30°，
∵DE=1，
∴AE=2DE=2，
∵∠FAE=∠AFD=30°，
∴EF=AE=2，
Therefore, B
As shown in the figure, in RT △ ABC, ∠ C = 90 ° and the vertical line ed of AB intersects BC with D, and ∠ CAD: Cab = 13 And ∠ CAD: Cab = 1:3, calculate the degree of ∠ B~~ Everybody, help ha ~ ~ picture, ha ha, temporarily mi you ~ ~ is the right triangle (1) extracurricular exercise problem 7 ~ ~ 55 little sister, thank you again~~
Because ED is the perpendicular of ab
So ad = BD
So ∠ DAB = ∠ B
Because ∠ CAD: Cab = 1:3
So ∠ cab = 3 ∠ CAD ∠ B = ∠ DAB = 3 ∠ CAD - ∠ CAD = 2 ∠ CAD
Because ∠ B + ∠ cab = 90 °
So 3 ∠ CAD + 2 ∠ CAD = 90 degrees
The result shows that CAD = 18 degrees
∠B=2∠CAD=36°
As shown in the figure, in △ ABC, ∠ C = 90 °, the vertical bisector De of segment AB intersects BC with D, and the perpendicular foot is e. if ∠ cab = 65 °, then ∠ CAD=______ °．
In ∵ ABC, ∵ C = 90 ﹤ cab = 65 °,
∴∠B=180°-∠C-∠CAB=180°-90°-65°=25°．
∵ the vertical bisector De of line AB intersects BC at D,
∴AD=DB，∠DAB=∠B=25°．
∴∠CAD=∠CAB-∠DAB=65°-25°=40°．
It is known that in △ ABC, the bisector ad of ∠ cab and the vertical bisector De of BC intersect at point D, DM ⊥ AB and m, DN ⊥ AC intersect the extension of AC at n. what do you think is the relationship between BM and CN? Try to prove your findings
BM=CN．
Reason: connect BD, CD,
∵ ad bisection ∵ BAC, DM ⊥ AB, DN ⊥ AC,
∴DM=DN，
∵ de bisects BC vertically,
∴BD=CD，
In RT △ BMD and RT △ CND
A kind of
BD＝CD
DM＝DN
∴Rt△BDM≌Rt△CDN（HL），
∴BM=CN．
As shown in the figure, ∠ cab = 90 ° in △ ABC, the vertical bisector of CB intersects BC at point E, and the extension of CB intersects Ca at point D The result showed that AE ^ 2 = EF · ed
∵∠BAC=90°,∴∠B+∠C=90°,
∵FE⊥BC,BE=CE,
∴∠F+∠C=90°,AE=BE,
Ψ B = ∠ f = ∠ DAE, and ∠ CEF is the common angle,
∴ΔEAD∽ΔEFA,
∴DE/AE=AE/EF,
∴AE^2=EF*ED.
As shown in the figure, it is known that in the triangle ABC, the angle ACB = 90 °, the points D and E are all on AB, and ad = AC, the angle DCE = 45 °, indicating BC = be Please.
It is proved that: the following proof is: the ? ACB = 90 ? a + ∵ B = 90 ? B = 90 ? a ? ad = AC ∵ a ∵ ad = AC ∵ AC ? ad = AC ∵ ADC ? ad = AC ? a ? a ? a ? a ? ad = AC = AC ? ad = AC A / 2 = 45 +
As shown in the figure, in the right triangle ABC, the angle ACB is equal to 90 degrees, D and E are the points on AB, and ad is equal to AC, be is equal to BC, and the angle DCE is calculated Can't give the picture, 2.5 right triangle (1)
In triangle ECD
Because the angle ECD = 180 - (angle CED + angle CDE)
And ad = AC, be = BC
So angle ADC = angle ACD, angle CEB = angle ECB
So the angle ECD = 180 - (angle ECB + angle ACD)
Because the angle ACB is 90
That is, angle ACD + angle ECB - angle ECD = 90 (because angle ECD is added once more)
Therefore, 180 angle ECB angle ACD = 90 angle ACD angle BCE + 2 angle ECD
Both 180 = 90 + 2 angle ECD
2 angle ECD = 90
Angle ECD = 45
In △ ABC, ∠ ACB = 90 °, points D and E are on AB, and ad = AC, BC = be. Find the degree of ∠ DCE
∵AD=AC，BC=BE，
∴∠ACD=∠ADC，∠BCE=∠BEC，
∴∠ACD=（180°-∠A）÷2①，∠BCE=（180°-∠B）÷2②，
∵∠A+∠B=90°，
ν ① + ② - ∠ DCE, ∠ ACD + ∠ BCE - ∠ DCE = 180 ° - (∠ a + ∠ b) ÷ 2 - ∠ DCE = 180 ° - 45 ° - ∠ DCE = 135 ° - ∠ DCE = 90 °,
∴∠DCE=45°．
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