In RT △ ABC, ∠ a = 90 °, ab = AC, D is the midpoint of BC, e and F are on AC and AB, and AE = BF. It is proved that △ DEF is an isosceles right triangle

In RT △ ABC, ∠ a = 90 °, ab = AC, D is the midpoint of BC, e and F are on AC and AB, and AE = BF. It is proved that △ DEF is an isosceles right triangle

Connect ad, AC = ab
AD ⊥BC,∠DAC=∠B=45°
AD=BD,Ae=BF
△BDF≌△ADe
DF=DE
∠BDF=∠ADE
∠BDF+∠FDA=90°
∠ADE+∠FDA=90°
Δ DEF is an isosceles right triangle

In the isosceles triangle ABC, ab = AC, e, f are points on AB and AC respectively, AE = CF, BF and CE intersect at point D, and D is the midpoint of BF

Because AB = AC, AE = CF, so AB AE = AC CF, that is, be = AF, so AE / AF = AE / be, because AB is parallel to FH, so AE / be = FG / GH, because be is parallel to FG, BD = FD, so bed and FGD are congruent triangles, so FG = be, so FG / GH = be / GH

2. In △ ABC, ab = CB, D is a point on BC edge, e is a point on ad edge, and AC ^ 2 is satisfied= CD.CB ,AE/BD=AC/AB （1）CD=CE 2. In △ ABC, ab = CB, D is a point on BC edge, e is a point on ad edge, and AC ^ 2 is satisfied= CD.CB , AE / BD = AC / AB (1) CD = Ce (2) de / AC = DC / AB (3) when BD = CD, find the value of s △ CDE: s △ CAE

1. ∵ AB = CB  BAC = ∠ BCA in △ ABC and △ ACD, AC] = CD × CB, that is, AC / BC = CD / AC ∠ ACB = ∠ ACD (in the same angle) ｛ ABC ∵ ACD

The results show that △ ABC is isosceles triangle, △ BDC and △ ace are isosceles triangle respectively. AE and BD intersect at point F, connect CF and extend, intersect AB at point G The midpoint of ab

prove:
∵AC＝BC
∴∠CAB＝∠CBA
∵ equilateral △ BDC, equilateral △ ace
∴∠CBD＝∠CAE＝60
∵∠BAE＝∠CAB-∠CAE,∠ABD＝∠CBA-∠CBD
∴∠BAE＝∠ABD
∴AF＝BF
∵CF＝CF
∴△ACF≌△BCF （SSS）
∴∠ACG＝∠BCG
/ / Ag = BG (three wires in one)
/ / G is the midpoint of ab

As shown in the figure, in the triangle ABC, BD and CD are the bisectors of angle ABC and angle ace respectively, and BD and CD intersect at point D. It is proved that the angle BDC = half angle a = 90 degrees

In the triangle ABC, BD and CD are the bisectors of ∠ ABC, ∠ ACB, and BD and CD intersect at point D. It is proved that ∠ BDC = (1 / 2 ∠ a + 90) degree

As shown in the figure, BD is the bisector of the angle of triangle ABC, and CD is the bisector of the outer angle ace of triangle ABC. They intersect at point D. try to explore the angle between angle BDC and angle A Quantitative relation

prove:
∵∠A+∠ABC+∠ACB＝180
∴∠ABC+∠ACB＝180-∠A
∵ ace = 180 - ∠ ACB, CD bisection ∠ ace
∴∠DCE＝∠ACE/2＝（180-∠ACB）/2＝90-∠ACB/2
∵ BD bisection ∵ ABC
∴∠DBC＝∠ABC/2
∵ DCE is the external angle of  DBC
∴∠DCE＝∠D+∠DBC＝∠D+∠ABC/2
∴∠D+∠ABC/2＝90-∠ACB/2
∴∠D＝90-（∠ABC+∠ACB）/2＝90-（180-∠A）/2＝∠A/2
This is a similar topic I did a few days ago, please refer to

As shown in the figure, BD is the bisector of the angle of triangle ABC, and CD is the bisector of ace of the outer angle of triangle ABC. Find the quantitative relationship between angle BDC and angle A

∵ CD bisection ∵ ace
∴∠ACD=∠ECD
∵∠ECD=∠CBD+∠D
∴2∠ECD=2∠CBD+2∠D
∵∠ABD=∠CBD
∴∠ACE=∠ABC+2∠D
∵∠ACE=∠ABC+∠A
∴∠A=2∠D

In RT △ ABC, ∠ ACB = 90 ° and isosceles triangle abd and ace are respectively made from AB and AC to the outside of the triangle ABC And ad is perpendicular to AC, AE is perpendicular to AB, connecting de and crossing AB to point F Xiao Ming thinks in this way: as shown in Figure 14, when ∠ BAC = 45 °, make eg ⊥ AC cross AB at point G, then FA = FG Xiaoying thinks like this: as shown in Figure 15, when ∠ BAC = 30 ° is, do DG ‖ AE and cross AB to point G, then FA = FG (1) Are Xiaoming and Xiaoying correct? Please explain the reasons (2) Please select one of the two graphs to explore the quantitative relationship between the segments FB and FA, and explain the reasons

(1) Xiao Ming's judgment "Fa = FG" is correct. (see left figure) it is proved that: ? BAC = 45 °, ad ⊥ AC, AE ⊥ ab.  DAB = ∠ EAC = 45 °; Da = dB, EA = EC. ⊿ AEC and ⊿ ADB are isosceles right triangle, quadrilateral acbd is square

As shown in the figure, in △ ABC, ab = AC, ∠ BAC = 40 ° respectively, make two isosceles right angle triangles abd and ace with AB and AC as the sides, so that ∠ bad = ∠ CAE = 90 ° (1) Find the degree of ∠ DBC; (2) Confirmation: BD = CE

(1)  abd is an isosceles right triangle,  DBA = 45 ° and ? AB = AC, ∵ BAC = 40 ° and  ABC = 70 °.  DBC = 115 °; (2) it is proved that ? abd and △ ace are isosceles right triangle, ? bad = ∠ CAE = 90 °, ab = ad, AC = AE

Given the triangle ABC, take AB and AC as the sides, and make the triangle abd and triangle ace on the outside of the triangle ABC respectively, so that ab = ad, AC = AE, bad = angle EAC, When the angle bad = 90 °, if the angle BAC = 45 °, angle BAP = 30 °, BD = 2, find the length of CD

∵∠BAD=∠EAC=90°
∴∠BAD+∠BAC=∠BAC+∠EAC
That is ∠ DAC = ∠ BAE
∵ ad = AB = √ 2 / 2 × 2 = √ 2 (using Pythagorean theorem)
AC=AE
∴△ACD≌△ABE
∴∠ADC=∠ABC
∠AEB=∠ACD
The four points a, D, B and P are circular. A, P, C and E are circular
∴∠BAP=∠BDP=30°
∠BPD=∠BAD=90°
/ / in RT △ BDP
BP=1/2BD=1
∴PD=√(BD²-PB²)=√(2²-1²)=√3
∵∠ABP=90°-∠BDP-∠DBA=90°-30°-45°=15°
In △ ABP, sine theorem: AP/sin15 ° =BP/sin30 °
AP=BP×sin15°/sin30°=2sin15°
∵∠PAC=∠BAC-∠BAP=45°-30°=15°
Ψ BEC = ∠ PAC = 15 ° (the four points above are circular)
∴∠AEB=∠ACP=∠AEC-∠BEC=45°-15°=30°
In △ ACP
AP/sin30°=PC/sin15°
PC=AP×sin15°/sin30°=4sin²15=4×(√6-√2)²/16=2-√3
∴CD=PD+PC=√3+2-√3=2