As shown in the figure, ad is the bisector of ∠ BAC in △ ABC, P is any point on ad, and ab > AC

Prove: as shown in the figure, cut AE on AB, make AE = AC, connect PE, ∵ ad is the bisector of ∠ BAC, ∵ bad = ∠ CAD, in △ AEP and △ ACP, AE = AC ∠ bad = ∠ cadap = AP, ≌ AEP ≌ △ ACP (SAS), ≌ PE = PC, in △ PBE, be > pb-pe, namely AB-AC > pb-pc

In triangle ABC, angle ACB = 90 degrees, AC = BC, P is a point in the triangle, PA = 3, BP = 1, PC = 2, find the degree of angle BPC

Fix point C of triangle BCP and rotate the rest until CB and Ca coincide
Suppose that the current position of the original point P is D, then:
Obviously, the rotated graph (triangle CAD) and the original graph (triangle CBP) must be identical
CD=CP=2,AD=BP=1,
Angle BCP = angle ACD. Angle BPC = angle ADC
So angle DCP = angle ACB - angle BCP + angle ACD = 90 degrees
So if we connect DP, then:
In triangle DPC, according to Pythagorean theorem, we get DP = 2 * radical 2
So in the triangle ADP,
AD*AD+DP*DP=AP*AP.
So triangle ADP is a right triangle, angle ADP = 90 degrees
According to the above analysis, only the angle ADC is required for BPC
Angle ADC = angle ADP + angle CDP
The former is a right angle, and the latter is a base angle of an isosceles right triangle CDP, equal to 45 degrees
So the angle ADC is 135 degrees
So we get the degree of BPC

It is known that in the triangle ABC, the angle ACB = 90 degrees, AC = BC, P is a point in the triangle ABC, and PA = 3, BP = 1, PC = 2, find the degree of the angle BPC

The answer is 135 degrees. Rotate the triangle ACP around point C to make AC and BC coincide, and connect the new p with the original P, that is, PC New P triangle is isosceles right angle, P NEW PB triangle is right angle, 45 + 90 is 135

As shown in the figure, BP and CP are bisectors of ∠ CBD and ∠ BCE at the outer corner of ABC respectively

It is proved that PM ⊥ ad at point m, PN ⊥ BC at point n, PG ⊥ AC at point G through point P, ∵ BP and CP are the bisectors of ∠ CBD and ∠ BCE at the outer angles of ABC, respectively, ∵ PM = PN, PG = PN, ∵ PM = PG, ∵ P is on the bisector of ∠ BAC

Am is the middle line of triangle ABC, D is a point in BM, de / / am is called AB, CA extension line and E, F, prove de + DF = 2am

In △ ABM, because de / AM has de / am = BD / BM. (1) in △ CED, by AM / / DF, there is am / DF = cm / CD, and conversely there is de / am = CD / cm. (2) because am is the middle line of △ ABC, M is the middle point of BC, and BM = cm = (1 / 2) BC, then (1) + (2) formula (DF + de) / A

In △ ABC, if AB = 3, O is the outer center of △ ABC, and the vector OA * vector BC = 1, then AC=

Let the midpoint of BC be D, then BC is perpendicular to OD by the property of the outer center
So 1 = OA * BC = (OD + DC + Ca) * BC = (BC / 2 + Ca) BC = (BA / 2 + AC / 2 + Ca) (Ba + AC)
=1/2(BA-AC)(BA+AC)=1/2(BA^2-AC^2)
So the length of AC is root 7

In the right triangle ABC, ∠ BAC = 90 °, BD bisects ∠ ABC, intersects AC at point D, passes through point C as CE ⊥ BD, intersects the extension line of BD at point E, intersects the extension line of BA at point F
Connecting DF. Proving BD = CF

It is proved that when Ba and CE are extended, the two lines intersect at point F
∵BE⊥CE
∴∠BEF=∠BEC=90°
In △ bef and △ bec
∠FBE=∠CBE, BE=BE, ∠BEF=∠BEC
∴△BEF≌△BEC(ASA)
∴EF=EC
∴CF=2CE
∵∠ABD+∠ADB=90°,∠ACF+∠CDE=90°
And ∵ ∠ ADB = ∠ CDE
∴∠ABD=∠ACF
In △ abd and △ ACF
∠ABD=∠ACF, AB=AC, ∠BAD=∠CAF=90°
∴△ABD≌△ACF(ASA)
∴BD=CF
∴BD=2CE

As shown in the figure, it is known that in RT △ ABC, ab = AC, BD bisects ∠ ABC, CE ⊥ BD intersects BD extension line at e, Ba and CE extension line intersects at F. it is proved that: (1) Δ BCF is isosceles triangle; (2) BD = 2ce

It is proved that: (1) BD bisects ∵ ABC, ∵ FBE = ∵ CBE. ∵ CE ⊥ BD, ∵ bef = ∵ BEC = 90 ° and ∵ be = be, ≌ bef ≌ BEC, ≌ BF = BC, that is ? BCF isosceles triangle. (2) ? BF = BC, CE ⊥ BD, ≁ CF = 2ce = 2ef, ? abd + ≌ ADB = 90 °, abd + ≌ AFE = 90 °

It is known that, as shown in the figure, ad and be are the heights of △ ABC, the extension lines of AD and EB intersect at h, and BH = AC

Intercept HF = BC on DH and connect BF
He vertical BC, CD vertical ah
Angle CBE = angle DBH
So angle c = angle h,
AC=BH,BC=HF
ABC congruent BFH
BF = AB, HbF = BAE
Angle BAE + angle Abe = 90 degree
So angle HbF + angle Abe = 90
So the angle ABF is 90 degrees
So the triangle ABF is an isosceles right triangle,
BD bisector Abf
Angle abd = 45 degrees
So the angle ABC = 135 degrees
Are you from the high school attached to North Jiaotong University? The first problem of our math homework today is this!

In RT △ ABC, if the bisector of acute angle ∠ ABC and ∠ cab intersects at point O, then ∠ boa=______ ．

As shown in the figure, in RT △ ABC, ∠ cab + ∠ ABC = 90 °, ∵ AO and Bo are the bisectors of ∠ cab and ∠ ABC, respectively, ∠ 1 = 12 ∠ cab, ∠ 2 = 12 ∠ ABC, ∠ 1 + ∠ 2 = 12 (∠ cab + ∠ ABC) = 12 × 90 ° = 45 °, in △ AOB, ∠ boa = 180 ° - (∠ 1 + ∠ 2) = 180 ° - 45 ° = 135 °

As shown in the figure, ad is the bisector of ∠ BAC in △ ABC, P is any point on ad, and ab > AC