As shown in the graph, △ ABC, ab = AC, points D, e, f are on edges BC, AB, AC respectively, and BD = CF, ∠ EDF = ∠ B, is there any triangle congruent with △ BDE in the graph? And explain the reason

As shown in the graph, △ ABC, ab = AC, points D, e, f are on edges BC, AB, AC respectively, and BD = CF, ∠ EDF = ∠ B, is there any triangle congruent with △ BDE in the graph? And explain the reason

Existence, △ BDE ≌ △ CFD
Reason: ? EDC = ∠ EDF + ∠ CDF,  EDC = ∠ B + ∠ bed,
∴∠EDF+∠CDF=∠B+∠BED，
And ? EDF = ∠ B,
∴∠BED=∠CDF．
∵AB=AC
∴∠B=∠C
∵BD=CF
∴△BDE≌△CFD（AAS）．

As shown in the figure, in △ ABC, ab = AC, D, e, f are on AB, BC, AC respectively, and BD = CE, ∠ def = ∠ B, is there any triangle congruent with △ BDE

Existence, △ CEF

As shown in the figure, in the triangle ABC, ad is high, AE and BF are angular bisectors. They intersect at point O, angle a is 60 ° and angle c is 50 ° to find angle DAC and angle boa Yes

In the triangle DAC

In the triangle ABC, ad is the center line on BC side and F is any point on ab. CF intersects ad with e. it is proved that AE * BF = 2DE * AF

Extend ed to G so that DG = De, connect BG,
Because BD = CD
So BG / / CE
EF / / BG
So AF / BF = AE / eg
AE*BF=EG*AF
EG=2DE
So AE * BF = 2DE * AF

As shown in the figure, in the triangle ABC, ab = 8, AC = 6, then the value range of the midline ad on the BC side

Extend ad to point m so that DM = ad, and connect BM. First, prove that triangle ADC and triangle BDM are congruent, AC = BM. Then in triangle AMB, am is greater than AB + BM, am is less than ab-bm, so the range of AM is obtained. Do you know?

As shown in the figure, in △ ABC, ab = 17cm, BC = 16cm, midline ad = 15cm on BC side, is △ ABC an isosceles triangle? Why?

Delta ABC is an isosceles triangle,
∵ ad is the center line of BC side, BC = 16cm,
∴BD=DC=8cm，
∵AD2+BD2=152+82=172=AB2，
∴∠ADB=90°，
∴∠ADC=90°，
In RT △ ADC,
AC=
AD2+DC2=
152+82=17cm．
ν AC = AB, that is, △ ABC is an isosceles triangle

As shown in the figure, in △ ABC, the point P is a point in △ ABC. Try to prove that: ∠ BPC = ∠ a + ∠ ABP + ∠ ACP

It is proved that as shown in the figure, the intersection of BP and AC is extended at point D,
In △ abd, ∠ 1 = ∠ a + ∠ ABP,
In △ CPD, ∠ BPC = ∠ 1 + ∠ ACP,
∴∠BPC=∠A+∠ABP+∠ACP．

It is known that P is a point on the plane where △ ABC is located. It makes △ ABP, △ BCP and △ ACP all isosceles triangle. How many points of P satisfy the condition? Have a picture

You make the three corners of a triangle bisector, and the intersection point is the center point of the triangle
Moreover, the three newly formed triangles are isosceles triangles
Only this point meets the requirements of the topic

As shown in the figure, point P is a point in △ ABC. Try to explain: ∠ BPC = ∠ a + ∠ ABP + ∠ ACP

Angle BPC = 180 - ∠ PBC - ∠ PCB
180=∠A+∠ABP+∠ACP+∠PBC+∠PCB
Therefore: ∠ BPC = ∠ a + ∠ ABP + ∠ ACP

As shown in the figure, △ ABC is an isosceles right triangle, BC is a hypotenuse. When △ ABP is rotated anticlockwise around point a, it can coincide with △ ACP ′. If AP = 3, then the length of PP ′ is equal to () A. 3 Two B. 2 Three C. 4 Two D. 3 Three

According to the properties of rotation, it is easy to obtain △ ACP ′≌ △ ABP, ∠ BAP = ∠ cap ′, AP = AP ′,
∵∠BAP+∠PAC=90°，
∴∠PP′C+∠PAC=90°，
△ app 'is an isosceles right triangle,
From Pythagorean theorem to obtain PP '=
AP2+AP′2=
32+32=3
2．
Therefore, a