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Given that the exterior angles CBE and BCF of DB and DC bisecting triangle ABC intersect at point D, the relationship between angle D and angle a is proved?
∠D=90°-1/2∠A
The idea of proof: extend Ba to get the outer angle ∠ GAC adjacent to ∠ a, and ∠ GAC = 180 ° - ∠ BAC
From the angle of polygon and 360 degree, we can see that ∠ GAC + ∠ CBE + ∠ BCF = 360 degree
It is concluded that ∠ CBE + ∠ BCF = 360 ° - GAC
Because dB and DC share ∠ CBE and ∠ BCF equally
Therefore, DBC + DCB = 1 / 2 (CBE + BCF) = 1 / 2 (360 ° - GAC) = 180 ° - 1 / 2 ∠ GAC)
From ∠ GAC = 180 ° - bac
It is concluded that ∠ DBC + ∠ DCB = 180 ° - 1 / 2 ∠ GAC = 180 ° - 1 / 2 (180 ° - Bache) = 90 ° + 1 / 2 Bache
So ∠ d = 180 ° - (∠ DBC + ∠ DCB) = 180 ° - (90 ° + 1 / 2 ∠ BAC) = 90 ° - 1 / 2 ∠ a
As shown in the figure, in the triangle ABC, the bisectors of the inner angle a, the outer angle CBE and the angle BCF intersect at P, AP intersects BC at D, and BG ⊥ AP intersects g through B
1) If the angle GBP = 45 °, verify AC ⊥ BC
2) Make the high DH of triangle PDC on the side of PC, and explore the quantitative relationship between angle APB and angle HDC to explain the reason
1) As ∠ GBP = 45 ° and ∠ BGP = 90 °, so ∠ BPG = ∠ GBP = 45 ° and BP bisects ∠ CBE, so ∠ EBP = ∠ CBP, ∠ EBP = ∠ BAP + ∠ APB, ∠ CBP = ∠ GBP + ∠ DBG, because ∠ APB = ∠ GBP, ∠ BAP = ∠ DBG
If △ ABC and △ def can coincide completely, and the circumference of △ ABC is 80cm, de = 30 cm.DF=25cm , then BC=____ CM
25cm or 30cm
If △ ABC ≌ △ def, although the circumference of △ ABC is 100cm, a and B correspond to D and e respectively, and ab = 30cm, DF = 25cm, then the length of BC is?
AB=DE=30 DF=AC=25 BC=100-30-25=45
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