As shown in the figure, ∠ ACB = 90 ° in △ ABC, make isosceles triangle △ ABC with AC as bottom edge, ad = CD = 10, make de ⊥ AC through point D, perpendicular foot is f, de and ab intersect at point E, connect CE. (1) prove: AE = CE = be; (2) if AB = 15cm, BC = 9cm, P is a point on ray De, then when DP is the value, the perimeter of △ PBC is the smallest, and calculate the perimeter of △ PBC at this time

As shown in the figure, ∠ ACB = 90 ° in △ ABC, make isosceles triangle △ ABC with AC as bottom edge, ad = CD = 10, make de ⊥ AC through point D, perpendicular foot is f, de and ab intersect at point E, connect CE. (1) prove: AE = CE = be; (2) if AB = 15cm, BC = 9cm, P is a point on ray De, then when DP is the value, the perimeter of △ PBC is the smallest, and calculate the perimeter of △ PBC at this time

(1) It is proved that: ∵ de ⊥ AC, ∵ ACB = 90 °, ∵ EF ∥ BC, and ∵ ADC are isosceles triangles, ∵ point F is the midpoint of AC (the property of three lines in one of isosceles triangles), ∵ EF is the median line of △ ABC, that is to say, point E is the midpoint of hypotenuse AB, ∵ in RT △ ABC, AE = CE = be; (2) ∵ ACB = 90 °, ab = 15cm, BC = 9cm, ∵ AC = AB2 − BC2 = 152 − 92 = 12, ∵ D = CD = 10cm, de ⊥ AC, f is the middle point of AC, EF = 12bc = 12 × 9 = 4.5, AF = 12ac = 12 × 12 = 6, DF = ad2 − af2 = 102 − 62 = 8, de = DF + EF = 8 + 4.5 = 12.5cm. According to the knowledge of axisymmetric shortest path, when point P coincides with point E, Pb + PC is the smallest, that is, the perimeter of △ PBC is the smallest. At this time, Pb = PC = 12ab = 152, that is, when DP = de = 12.5cm, the perimeter of △ PBC is the smallest, and the perimeter of △ PBC is the smallest Perimeter = Pb + PC + BC = 15 + 9 = 24cm

If P is the midpoint of AC, find the tangent of the dihedral angle p-bd-c
Needless to say, can we elaborate on our thinking?

Make PE vertical BC in plane ABC, intersect BC with E, then PE = EC = PC * sin45 degree = (radical 2) / 2
And: BC = 2 (root 2), so: be = bc-ec = (3 / 2) (root 2), be / BC = 3 / 4
In plane abd, if the EF is perpendicular to BD and intersects BD with F, then the EF is parallel to CD, EF / CD = be / BC = 3 / 4
EF = (3 / 4) CD = (3 / 4) (root 6)
It can be proved that BD is vertical to PE
(let m be the midpoint of BC and n be the midpoint of BD, then: Mn is parallel to CD, BD is vertical to Mn; and abd is isosceles triangle, so BD is vertical to an; therefore, BD is vertical to amn, BD is vertical to an; and in triangle ABC, am is vertical to BC, am is parallel to PE, so BD is vertical to PE)
Therefore, the BD vertical plane PEF, dihedral angle p-bd-c = angle PFE
In triangle PBD, PF is vertical to BD, so pf = (PD ^ 2-fd ^ 2) ^ (1 / 2)
PD can be solved by triangle ADC (2,2, √ 6, P is the midpoint of AC); FD can be solved by triangle abd (FD / BD = EC / BC, BD = (BC ^ 2-CD ^ 2) ^ (1 / 2))
So, in the triangle PFE, we get PF, PE, Fe, then we can get cos (angle PFE) by using the coxuan theorem, so we can get sin (angle PFE), Tan (angle PFE)

As shown in the figure, in △ ABC, ad ⊥ BC, CE ⊥ AB, D and e respectively, ad and CE intersect at point h, please add an appropriate condition:______ So that △ AEH ≌ △ CEB

∵ ad ⊥ BC, CE ⊥ AB, D and E, respectively, ∠ BEC = ∠ AEC = 90 ° in RT △ AEH, ∵ eah = 90 ° - ∠ ahe, and ∵ eah = ∠ bad, ∵ bad = 90 ° - ∠ ahe, in RT △ AEH and RT △ CDH, ∵ CHD = ∠ ahe, ∵ eah = ∠ DCH, ∵ eah = 90 ° - ∠ CHD = ∠ BCE, the results are as follows

In the triangle ABC, ad is vertical to BC, CE is vertical to AB, and the perpendicular feet are D, ad and CE at the intersection point h. When the angle BAC is equal to what, the triangle AEH is equal to the triangle CEB

45 ° for the following reasons:
By ad ⊥ BC, CE ⊥ AB,
∴∠EAH+∠B=90°,
And ∠ ECB + ∠ B = 90 °,
∴∠EAD=∠ECB.
When AE = CE,
△EAH≌△ECB,（A,S,A）,
The ∧ AEC is an isosceles right triangle,
∴∠BAC=45°.

As shown in the figure, in △ ABC, ad ⊥ BC, CE ⊥ AB, D and e respectively, ad and CE intersect at point h, please add an appropriate condition:______ So that △ AEH ≌ △ CEB

∵ ad ⊥ BC, CE ⊥ AB, D and E, respectively, ∠ BEC = ∠ AEC = 90 ° in RT △ AEH, ∵ eah = 90 ° - ∠ ahe, and ∵ eah = ∠ bad, ∵ bad = 90 ° - ∠ ahe, in RT △ AEH and RT △ CDH, ∵ CHD = ∠ ahe, ∵ eah = ∠ DCH, ∵ eah = 90 ° - ∠ CHD = ∠ BCE, the results are as follows

As shown in the figure, in △ ABC, ab = AC, angle a = 120 °, the vertical bisector of AB intersects BC at m, AB at e, and the vertical bisector of AC intersects AC at F. the proof is BM = Mn = NC

It is proved that: ∵ AB = AC, ∵ a = 120 & # 186; ∵ B = ∵ C = (180 & # 186; - 120 & # 186;) △ 2 = 30 & # 186; connecting am, an ∵ me is the vertical bisector of ab ∵ BM = am ∵ B = ∵ mAb = 30 & # 186; ∵ NF is the vertical bisector of AC ∵ an = cn ∵ C = ∵ NAC = 30 & # 186; man = 120 & # 186; - mAb -

As shown in the figure, △ ABC, BC = 2Ac, ∠ DBC = ∠ ACB = 120 °, BD = BC, CD intersection AB at point E. (1) calculate the degree of ∠ ace. (2) prove: de = 3ce

(1) In this paper, we prove that if we pass through point B, we can make BM ⊥ DC

It is known that in △ ABC, BC = 2Ac, ∠ DBC = ∠ ACB, BD = BC, CD intersection AB is at point E. when the angle ACB = 120 °, de = 3ce is proved

∵∠DBC=∠ACB=120°,BD=BC
∴∠D=∠BCD=30
∴∠ACD=90
B as BM ⊥ DC to m, DM = MC.BM ＝1／2BC
∵AC=1/2BC
∴BM=AC
∵∠BMC＝∠ACM＝90
∠MEB＝∠CEA
∴BME≌ACE
∴ME＝CE＝1/2CE
DE=3CE

In △ ABC, BC = 2Ac, ∠ DBC = ∠ ACB = 120 ° BD = BC, CD intersects AB at point E, and the degree of ∠ ABC is calculated

Digression:
In △ ABC, BC = 2Ac, ∠ ACB = 120 °. With these two conditions, we can find the degree of ∠ ABC
Why is there so much nonsense in the middle
Take the midpoint m of BD and connect am
Because BC = 2Ac, CA = cm = MB
So △ AMC is an isosceles triangle with ∠ ACM as vertex angle and ∠ ACM = 120 degree
Therefore, CMA = cam = 30 degree
Let AC = 1, then CA = cm = MB = 1, then am = √ 3 (root 3, I'm afraid it won't show here)
Then in △ AMB, ∠ BMA = 150 °, MB = 1, am = √ 3 (root 3, I'm afraid it won't show here)
From the cosine theorem, ab = √ 7
Continue to use cosine theorem
∠ABC=∠ABM=arccos(5√7/14)

In triangle ABC, AC = BC, angle c = 20 degrees, D and E are points on edge BC and AC respectively. If angle CAD is equal to 20 degrees and angle CBE is equal to 30 degrees, calculate angle ade
We can only use congruence, not similarity

Make ∠ BAF = 20 ° in triangle cab, AF and CB intersect at F
Because CA = CB, ∠ C = 20 degree
Therefore, cab = CBA = ABF = 80 degree
Therefore, AFB = 80 degree
Therefore, AFB = ABF
So AB = AF
Therefore, EAF = 60 degree
Because ∠ EBA = 50 degree
Therefore, BEA = 50 degree
Therefore, EBA = bea
So AB = AE
So AE = AF
So delta AEF is an equilateral triangle
So AF = EF, ∠ AFE = 60 degree
Therefore, DFE = 40 degree
Because ∠ DAF = 60 ° - 20 ° = 40 ° and ∠ AFD = 100 °
Therefore, ADF = 40 degree
Therefore, ADF = DAF
So AF = DF
So DF = EF
Therefore, EDF = def = 70 degree
Therefore, ADE = 70 ° - 40 ° = 30 °
Jiangsu Wu Yunchao answers for reference!