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As shown in the figure, in △ ABC, ad is an angular bisector, e is a point on ad, and CE = CD. If AE = 2DE, the area of △ ace is 6, calculate the area of △ ABC
Because the area of △ ace is 6, AE = 2DE,
The area of △ CDE is 3
∵CE=CD
∴∠CDE=∠DEC
∴∠BDA=∠CEA
∵∠BAD=∠DAC
∴△ABD∽△CEA
/ / AE: ad = 2:3 is the similarity ratio
The area ratio is the square of the similarity ratio
The area of abd = 9 / 4 * 6 = 13.5
The area of △ ABC = 13.5 + 6 + 3 = 22.5
As shown in the figure, △ ABC is an equilateral triangle, extend BC to D, extend Ba to e, make AE = BD, connect EC and ED, try to explain CE = De
It is proved that: extend BD to f such that DF = BC, connect EF, ∵ AE = BD, △ ABC is an equilateral triangle,
As shown in the figure, △ ABC is an equilateral triangle, extend BC to D, extend Ba to e, make AE = BD, connect EC and ED, try to explain CE = De
Proof: extend BD to F, make DF = BC, connect EF,
∵ AE = BD, △ ABC is an equilateral triangle,
ν DF = BC = AB, that is, AE + AB = BD + DF, ∠ B = 60 °,
∴BE=BF，
BEF is an equilateral triangle,
∴∠F=60°，
In △ ECB and △ EDF,
BE＝EF
∠B＝∠F＝60°
BC＝DF ，
∴△ECB≌△EDF（SAS），
∴EC=ED．
As shown in the figure, △ ABC is an equilateral triangle, extend BC to D, extend Ba to e, make AE = BD, connect EC and ED, try to explain CE = De
Proof: extend BD to F, make DF = BC, connect EF,
∵ AE = BD, △ ABC is an equilateral triangle,
ν DF = BC = AB, that is, AE + AB = BD + DF, ∠ B = 60 °,
∴BE=BF，
BEF is an equilateral triangle,
∴∠F=60°，
In △ ECB and △ EDF,
BE＝EF
∠B＝∠F＝60°
BC＝DF ，
∴△ECB≌△EDF（SAS），
∴EC=ED．
As shown in the figure, △ ABC is an equilateral triangle, extend BC to D, extend Ba to e, make AE = BD, connect EC and ED, try to explain CE = De
Proof: extend BD to F, make DF = BC, connect EF,
∵ AE = BD, △ ABC is an equilateral triangle,
ν DF = BC = AB, that is, AE + AB = BD + DF, ∠ B = 60 °,
∴BE=BF，
BEF is an equilateral triangle,
∴∠F=60°，
In △ ECB and △ EDF,
BE＝EF
∠B＝∠F＝60°
BC＝DF ，
∴△ECB≌△EDF（SAS），
∴EC=ED．
As mentioned above, in the triangle ABC, ab = AC, D is the upper point of AB, and CD is the base edge of isosceles triangle CDE, connecting AE. If AE is parallel to BC, it is proved that ABC is similar to CDE
It is proved that: the over point E is EP ? AC is in P, EQ ? AB, the extension line of Ba is Q, the intersection point of AC and De is set as m ∵ AB = AC ∵ ABC ∵ ACB ? AE ? BC ? BC ? BC ? BC proof: it is proved that: the over point E is e for EP ? AC, Q ? ab ⊸ ab ? AB 8 ab CE = de
It is known that in the isosceles triangle △ ABC, ∠ BAC = 90 ° AB = AC, BD is the AC midline, AE ⊥ BD at F, intersection BC at e, link de. find: ﹣ ADB = ∠ CDE RT
Angle CDB = angle CDE + angle EDB = angle AEB = angle CEA + angle ace, triangle ADF is similar to triangle bef, triangle DEF is similar to triangle AFB, angle AEB = angle ADB, angle DEA = angle ADB, so angle ADB = angle CDE
As shown in the figure, ∠ ACB = 90 °, a = 45 ° in △ ABC, AC = AE, BC = BD, it is proved that △ CDE is an isosceles triangle
Because ∠ ACB = 90 ° and ∠ a = 45 ° in △ ABC, so △ ABC is an isosceles right triangle, so AB = AC, angle a = angle B
Because AC = AE = BC = BD, △ ace and △ BCD are congruent, CD = CE, so △ CDE is an isosceles triangle
As shown in the figure, △ ABC is an equilateral triangle, D is the midpoint of AB, taking CD as the edge, making the equilateral △ ECD upward, connecting AE, proving that △ ade is an isosceles triangle
It is proved that ∵ △ ABC is an equilateral triangle,
ν BC = AC, ∠ ACB = 60 °, (2 points)
It can be divided into three equal sides
∴∠ACB=∠DCE，
Ψ ACB - ∠ ACD = ∠ DCE - ∠ ACD, i.e., ∠ DCB = ∠ ace, (4 points)
In △ BDC and △ AEC,
BC＝AC
∠DCB＝∠ACE
CD＝CE ，
∴△BDC≌△AEC（SAS），
/ / BD = AE, (6 points)
∵ D is the midpoint of AB, ᙽ BD = ad,
∴AD=AE，
The △ ade is an isosceles triangle. (8 points)
As shown in the figure △ ABC and △ ECD are isosceles right triangles, point C is on ad, and the extension line of AE intersects BD at point F. verification: AF ⊥ BD
It is proved that ∵ is in △ ace and △ BCD
AC=BC
∠ACE=∠BCD=90°
CE=CD
∴△ACE≌△BCD，
∴∠CAE=∠CBD，
∵∠BCD=90°，
∴∠CBD+∠ADB=90°，
∴∠CAE+∠ADB=90°，
∴∠AFD=180°-90°=90°，
∴AF⊥BD．
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