As shown in the figure, in the parallelogram ABCD, the bisector of angle ab. C intersects with AD at point P. it is proved that PD + CD = BC

As shown in the figure, in the parallelogram ABCD, the bisector of angle ab. C intersects with AD at point P. it is proved that PD + CD = BC

Pg / / AB PD = GC PG = DC
Pg / / AB angle GPB = angle ABP = angle PBG
Δ PBG is an isosceles triangle PG = BG
PD＋DC＝GC＋PG＝GC＋BG＝BC

As shown in the figure, in the parallelogram ABCD, it is known that the bisector of ∠ bad intersects with edge BC at point E, bisector of ∠ ABC intersects with edge ad at point F, AE and BF intersect at point O. it is proved that abef of quadrilateral is diamond

The quadrilateral abef is a diamond,
∵AD∥BC,
∴∠1=∠2,
∵ BF bisection ∵ ABC,
∴∠2=∠3,
∴∠1=∠3,
∴AB=AF
Similarly, ab = be can be obtained,
∵AF∥BE,
The quadrilateral abef is a parallelogram,
∵AB=AF
The parallelogram abef is a diamond
So the answer is quadrilateral abef is diamond

Known: in triangle ABC, angle ACB = 2 angle B, ad vertical ab. verification: BD = 2Ac Where ∠ AB is perpendicular to a B. Confirmation: BD = 2Ac. AC is not vertical to DC, who can tell me how to paste it? And the triangle abd is a right triangle.

Let e be the midpoint of BD with Ae,
Then AE is the center line of the hypotenuse BD of the right triangle ADB
Therefore, AE = de = be = BD / 2
∠DEA=∠EAB+∠B=2∠B
Therefore, ∠ DEA = ∠ ACB
AC=AE
And: AE = BD / 2
So BD = 2Ac

As shown in the figure, in △ ABC, ∠ ACB = 2 ∠ B, BC = 2Ac

Make CD bisection ∠ ACB intersect AB at D, pass D as de ⊥ BC at E,
∵∠ACB=2∠B
Ψ B = ∠ BCD, that is, △ DBC is an isosceles triangle,
And de ⊥ BC,
ν BC = 2ce, and BC = 2Ac,
∴AC=EC，
∴△ACD≌△ECD
∴∠A=∠DEC=90°．

In this paper, we prove that △ B is a right angle of a triangle

D for de ⊥ AC
∵ ad is the angular bisector
∴BD/CD=AB/AC
∵AB=2AC
∴BD=2CD
Let CD = a, then BD = 2A
∵AD=BD
∴AD=2a
∵DE⊥AC
∴∠AED=90°,∠DAE=30°=∠BAD=∠B
Ψ C = 90 °, that is, C and D coincide
The △ ACB is a right triangle

In the triangle ABC, ∠ ACB = 2 ∠ B, BC = 2Ac Try to explain ∠ a = 90 degrees

Make the bisector CD of angle c and intersect AB with D;
Through D, De is perpendicular to BC and crossed BC to E
Because angle ACB = 2 angle B, so angle DCE = angle B, so triangle BCD is isosceles triangle
De is perpendicular to BC, so e is the midpoint of BC
For ACD and ECD triangles, there are AC = CE, ACD = angle ECD, DC = DC,
So triangle ACD and triangle ECD are congruent
So angle a is equal to angle Dec, and dec is a right angle
Get the certificate

As shown in the figure, △ ABC is connected to ⊙ o, AB is the diameter of ⊙ o, CD bisection ⊙ ACB intersects ⊙ o at point D, AB intersects AB at point F, chord AE ⊥ CD at point h, connecting CE and oh (1) Verification: △ ace ∽ CFB; (2) If AC = 6, BC = 4, find the length of oh

(1) It is proved that: ∵ AB is the diameter of ⊙ o,  ACB = 90; ∵ CD bisection  ACB,  ACD = ∵ FCB = 45 °; ? AE ⊥ CD, ? CAE = 45 ° = ∠ FCB; in △ ace and △ BCF, ? CAE = ∠ FCB, ? e = ∠ B, ∵ ace ∵ CFB; (2) extend the intersection point of AE and CB at M; ? FCB = 45 °

As shown in the figure, △ ABC is inscribed in ⊙ o, ad is the diameter of ⊙ o, and the cross chord BC is at point e. it is known that ∠ ACB = 60 ° and BC = 16cm (1) Find the degree of ∠ bad; (2) When ad ⊥ BC, find the diameter of ⊙ o

Connect BD,
(1) ∵ diameter ad,
∴∠ABD=90°，
∵∠C=60°，
∴∠BDA=60°，
∴∠BAD=30°，
（2）∵AD⊥BC，BC=16cm，
∴BE=CE=8cm，
∵∠BAD=30°，
∴AB=2BE=16cm，
∵∠ABD=90°，∠BAD=30°，
∴AD=32
Three
3cm．

It is known that in △ ABC, ab = AC, point D is on BC, and BD = ad, DC = AC, (1) Write two isosceles triangles; (2) Find the degree of ∠ B

（1）△ABC，△ACD．△ABD，
From ab = AC, △ ABC is an isosceles triangle, and from BD = ad, △ abd is an isosceles triangle;
From DC = AC, △ ACD is an isosceles triangle
(2) Let ∠ B = X,
∵BD=AD，
∴∠DAB=∠B=x，
∵AB=AC，
∴∠C=∠B=x，
∵DC=AC，
∴∠CAD=∠ADC=∠DAB+∠B=2x，
In △ ACD, from ∠ CAD + ∠ ADC + ∠ C = 180 °, 2x + 2x + x = 180,
It is found that x = 36 ° and ∠ B = 36 °
Answer: the degree of ∠ B is 36 degrees

In △ ABC, there are several isosceles triangles with ab = AC, ad = DC. ∠ BAC = 3 ∠ C

Three
∵AB=AC
ν isosceles △ ABC
∵ ad = CD ᙽ isosceles ᙽ ADC
∵∠BAC=3∠C ∠DAC=∠C ∠B=∠C
∴∠BAD=∠ADB=2∠C
ν isosceles △ abd