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In the triangle ABC, BD = DC = AC, D is the midpoint of BC and E is the midpoint of DC. It is proved that the bisection angle BAE of AD
AE is extended to G so that AE = eg, and DG is connected at the same time. From AE = eg, de = EC, angle c = angle CDG
AC=DG
From DC = AC, angle CDA = angle DAC, so angle ADB = angle DAC + angle c = angle ADC + angle CDG = angle ADG,
So the triangle ADB is equal to the triangle ADG, and the angle bad = angle GAD
In △ ABC, BD = DC = AC, e is the midpoint of DC
Extend AE to m, make em = AE, and connect DM
Easy to prove △ DEM ≌ △ CEA
∴∠C=∠MDE,DM=AC
BD = DC = AC
∴DM=BD,∠ADC=∠CAD
And ∠ ADB = ∠ C + ∠ CAD
∠ADM=∠MDE+∠ADC
∴∠ADM=∠ADB
∴△ADM ≌△ADB
∴∠BAD=∠MAD
That is, ad bisection ∠ BAE
In the isosceles triangle ABC, ab = AC, ∠ a = 20 °, D is a point on the edge of AB, and ad = BC, connect CD, and find ∠ BDC The question is right, if you can't do it
Taking AC as the edge, make a regular triangle ace, connected with de, CE, because AE = AC = AB, ∠ DAE = 60 + 20 = 80 = ∠ b.ad = BC, so the triangle ade is all equal to the triangle ABC, de = AC, ∠ Dec = 40, ∠ ACD = 10 ∠ EDB = 70, ∠ BDC = 30
In the isosceles triangle ABC, AB is equal to AC, angle a is equal to 20 degrees, D is the point above AB, and ad is equal to BC, connecting CD, then how many degrees is the angle BDC equal to
0
It is known that the triangle ABC and the triangle ECD in the figure are isosceles right triangles ∠ ACB = ∠ DCB = 90 degrees, D is a point on the edge of AB to prove BD = AE
In the title, there is a mistake in the known conditions, it should be: "isosceles right triangle ∠ ACB = ∠ DCE = 90 degrees", isn't it? Yes. As shown in the figure above: it is proved that in △ BCD and △ ace ? ACB = ∠ DCE = 90 ? ACB = ∠ ACD = ∠ DCE - ∠ ACD, that is ∠ DCB = ∠ ECA and ∵ CD = CE; BC = AC △ BCD
As shown in the figure, △ ABC and △ ECD are isosceles right triangles, ∠ ACB = ∠ DCE = 90 ° and D is a point on AB side (1) △ ace ≌ △ BCD, (2), ad ≌△ BCD, ad ≌ + AE ≌ = de
It is proved that: (1) the following: (1) the
As shown in the figure, triangle ABC and triangle ECD are isosceles right angle triangles, angle ACB equals angle DCE equals 90 degrees, and D is a point on ab
AC = BC, DC = EC
∵∠eca+∠acd=90
∠bcd+∠acd=90
∴∠eca=∠bcd
All △ ace is equal to BCD
∴bd=ae
As shown in the figure, in triangle ABC, ad is perpendicular to BC and D, and ad ^ 2 = BD * DC
It is proved that because ad ^ 2 = BD * DC, ad is perpendicular to BC, so triangle abd is similar to triangle CAD, so angle bad = angle C,
Because the angle c + angle CAD = 90 degrees, the angle BAC = 90 degrees, so the triangle ABC is a right triangle
As shown in the figure, △ ABC and △ DBE are isosceles right triangles (1) Results: ad = CE; (2) Ad CE and vertical conjecture? If vertical, please state the reason; if not, write the conclusion instead of the reason
(1) ∵ △ ABC and △ DBE are isosceles right triangle,
∴AB=BC，BD=BE，∠ABC=∠DBE=90°，
∴∠ABC-∠DBC=∠DBE-∠DBC，
Namely, abd = ∠ CBE,
∴△ABD≌△CBE，
∴AD=CE．
(2) The extended ad crossed BC and CE to G and f respectively,
∵△ABD≌△CBE，
∴∠BAD=∠BCE，
∵∠BAD+∠ABC+∠BGA=∠BCE+∠AFC+∠CGF=180°，
And ∵ BGA = ∵ CGF,
∴∠AFC=∠ABC=90°，
∴AD⊥CE．
As shown in the figure, △ ABC and △ DBE are isosceles right triangles (1) Results: ad = CE; (2) Confirmation: AD and CE are vertical
It is proved that: (1) ∵ ABC and △ DBE are isosceles right triangles,
∴AB=BC，BD=BE，∠ABC=∠DBE=90°，
∴∠ABC-∠DBC=∠DBE-∠DBC，
Namely, abd = ∠ CBE,
∴△ABD≌△CBE，
∴AD=CE．
(2) In prolonged ad, BC and CE were crossed to G and f respectively,
∵△ABD≌△CBE，
∴∠BAD=∠BCE，
∵∠BAD+∠ABC+∠BGA=∠BCE+∠AFC+∠CGF=180°，
And ∵ BGA = ∵ CGF,
∴∠AFC=∠ABC=90°，
∴AD⊥CE．
RELATED INFORMATIONS
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