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As shown in the figure, in △ ABC, ∠ abd = ∠ ACD = 60 ° and ∠ ADB = 90 ° - 1 2∠BDC． It is proved that △ ABC is an isosceles triangle
It is proved that: ∵ abd = ∵ ACD,
The four points a, B, C and D are circular,
∴∠ADB=∠ACB，∠BDC=∠BAC，
∵∠ADB=90°-1
2∠BDC，
∴∠ACB=90°-1
2∠BAC，
∴2∠ACB+∠BAC=180°
And ? ABC + ∠ ACB + ∠ BAC = 180
∴∠ABC=∠ACB，
∴AB=AC，
The △ ABC is an isosceles triangle
As shown in the figure, in △ ABC, ∠ abd = ∠ ACD = 60 ° and ∠ ADB = 90 ° - 1 2∠BDC． It is proved that △ ABC is an isosceles triangle
It is proved that: ? abd = ∠ ACD, a, B, C and D are in a circle, ? ADB = ∠ ACB, BDC = ∠ BAC, ADB = 90 ° - 12 ∠ BDC, ? ACB = 90 ° - 12 BAC, ? 2 ∠ ACB + ∠ BAC = 180 ° and ? ABC + ∠ ACB, AB = AC, △ ABC is equal to
In the triangle ABC, BC and CE are bisectors of angle ABC and angle ACB respectively. If angle abd = 20 ° and angle BDC = 80 ° then the degree of angle AEC is equal to
∵∠ABD=20º,∴∠DBC=20º,∠ABC=40º
In Δ BCD, ∠ ACB = 180-80-20 = 80 ° and ∠ ace = 40 °
In Δ ABC, ∠ a = 180-80-40 = 60 °,
In Δ ace, ∠ AEC = 180-60-40 = 80 °
It is known that, as shown in △ ABC, ab = AC, ∠ a = 90 °, the bisector CD of ∠ ACB intersects AB at point E, ∠ BDC = 90 ° to prove that CE = 2bd
In the case, the number of \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\flat
As shown in the figure, BD bisects ∠ ABC, Da ⊥ AB, ∠ 1 = 60 ° and ∠ BDC = 80 ° to find the degree of ∠ C
∵DA⊥AB，
∴∠A=90°．
∵ BD bisection ∵ ABC,
∴∠ABD=∠CBD=90°-∠1=90°-60°=30°．
∵∠BDC=80°，
∴∠C=180°-∠CBD-∠BDC=180°-30°-80°=70°．
As shown in the figure, BD bisects ∠ ABC, Da ⊥ AB, ∠ 1 = 60 ° and ∠ BDC = 80 ° to find the degree of ∠ C
∵DA⊥AB，
∴∠A=90°．
∵ BD bisection ∵ ABC,
∴∠ABD=∠CBD=90°-∠1=90°-60°=30°．
∵∠BDC=80°，
∴∠C=180°-∠CBD-∠BDC=180°-30°-80°=70°．
As shown in the figure, △ ABC, ab = AC, BD is the bisector of ∠ ABC, and ∠ BDC = 75 °, find the degree of ∠ BAC
∵ BD is the bisector of ∵ ABC
∴∠ABD=∠DBC
∵AB=AC，
∴∠ABC=∠ACB=2∠DBC
∵∠DBC+∠ACB+∠BDC=180°，∠BDC=75°，
∴3∠DBC+75°=180°
∴∠DBC=35°
∴∠BAC=75°-35°=40°
In the RT triangle ABC, ab = AC = 1, ad is the height on the hypotenuse BC. Take ad as the crease, fold up the triangle to make the angle BDC right angle, and prove that the plane abd is vertical Plane abd vertical plane BDC
Because AB = AC
So the height ad of the right triangle is perpendicular to BC
Because BD is perpendicular to DC, BD is perpendicular to AD
So BD is perpendicular to the plane ADC
Because BD is a straight line on the surface abd
So the plane abd, the vertical plane BDC
In the plane rectangular coordinate system, the coordinate of point a is the following sign 3-radical 2, the coordinate of point C is negative root 3-radical 2, point B is on the Y axis and s triangle ABC = root 3
AC = heel 3-root 2 - (negative root 3-root 2) = 2 root 3
Because s triangle ABC = root 3
That is ob * AC = root 3 * 2 = 2 root 3
Because AC = 2 root sign 3
So ob = 1
If the side length of the equilateral triangle ABC is a, ad is vertical BC, and D is a perpendicular foot. Fold the triangle abd along ad so that the angle BDC = 90 degrees, then the distance from the folded back point B to the straight line AC is?
Do AC vertical DP through D
Suppose that point B is folded to the position B '
Angle b'dc = 90 degrees
So db 'vertical plane ABC
Then b'p is vertical AC
B'p is the distance to the straight line AC
DP=DC*sin60=√3a/4,DB'=a/2
B’P^2=DP^2+DB'^2=7a^2/16
B'P=√7a/4
The distance from the folded back point B to the straight line AC √ 7a / 4
RELATED INFORMATIONS
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