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It is known that in isosceles RT △ ABC, ∠ a = 90 ° D is the midpoint of BC, e and F are points on AB and AC respectively, and EA = CF is satisfied. Verification: de = DF
Proof: connect ad, as shown in the figure,
∵ △ ABC is an isosceles right triangle, D is the midpoint of BC,
ν ad = DC, ad bisection ∠ BAC, ∠ C = 45 °,
∴∠EAD=∠C=45°，
In △ ade and △ CDF
EA＝CF
∠EAD＝∠C
AD＝CD ，
∴△ADE≌△CDF，
∴DE=DF．
D. Let BC = a, AC = B, ab = C, if the angle BAC = 90 ° and the area of triangle ABC is s, it is proved that s = AE * BD
Let AE be x and BD be y
According to the fact that the circumference of the triangle ADB is equal to that of the triangle ACD
c＋Y+AD=a－Y＋AD＋b
So y = (a + B-C) / 2
Similarly, x = (a + C-B) / 2 can be obtained
therefore
Xy = [(a + B-C) / 2] * [(a + C-B) / 2] (Simplified)
ab/2
And because
S= ab/2
So the proposition is proved
If the area of triangle ACD is the ratio of the area of triangle BCD to the area of triangle ACB, find the value of sina
∵ CQ = 1 / 3ce, namely CQ / CE = 1 / 3
ν CQ / EQ = 1 / 2, namely EQ / CE = 2
∵ E and F are the midpoint of AB and AC respectively
∴EF∥BC,
Prolonged BQ to ef to H,
∴∠PHB=∠CBQ
∵ BQ bisection ∵ CBP
∴∠CBQ=∠PBQ=∠PHB
∴BP=PH
∵EF∥BC
∴△BCQ∽△EHQ
EH/BC=EQ/CQ=2
∴EH=2BC=12
∵EH=PE+PH=PE+BP
∴PE+BP=12
It is known that in △ ABC, ch is the bisector of ∠ ACD and BH is the bisector of ∠ ABC It is proved that: ∠ a = 2 ∠ H
It is proved that ∵ ACD is an external angle of ABC,
∴∠ACD=∠ABC+∠A，
∵ 2 is an external angle of BCH,
∴∠2=∠1+∠H，
∵ ch is the bisector of ∵ ACD and BH is the bisector of ∵ ABC,
∴∠1=1
2∠ABC，∠2=1
2∠ACD，
∠ a = ∠ ACD - ∠ ABC = 2 (∠ 2 - ∠ 1), and ∠ H = ∠ 2 - ∠ 1,
∴∠A=2∠H．
In the triangle ABC, it is known that the angle ACB is a right angle and CD is the height on the hypotenuse ab. it is proved that triangle ACD ∽ triangle CBD ∽ triangle ABC
Oh! This thing is very simple. The conclusion is a famous law called projective theorem. You can search the proof of projective theorem and try it!
In the triangle ABC, D is a point in the triangle. It is proved that the angle CBD = angle a + angle ACD + angle abd
Connect AD and extend BC to E
∵∠BDE=∠BAE+∠ABD ∠CDE=∠CAE+∠ACD
∴∠BDE=∠BAE+∠ABD+∠CAE+∠CDE
=∠A+∠ABD+∠ACD
In △ ABC, if the three sides a, B and C are in equal proportion sequence, find the sine value of the minimum angle
b^2=ac,
a^2+b^2=c^2
a^2+ac=c^2
a^2+ac-c^2=0
The relationship between a and C can be found by finding the two roots of the above equation
The sine values of the three inner angles of the right angle △ ABC form an equal proportion sequence. If the smallest acute angle is angle a, then Sina=
sin90=1
sinA:sinB=sinB=1
sinA=(sinB)^2
sinA=(cosA)^2
(sinA)^2+sinA=1
sinA=(1-√5)/2
Sina = (1 + √ 5) / 2 > 1 (omitted)
In the RT triangle ABC, the sine of acute angles a and B are the two roots of the equation x ^ 2 - (x / 2) - M = 0 Sorry, it's x ^ 2 - (x / 2) + M = 0
From Weida's theorem
sinA+sinB=1/2
sinA*sinB=m
Square formula (1) to get
1+2sinAsinB=1/4
Namely
1+2m=1/4
therefore
m=-3/8
Acute triangle function: in RT △ ABC, ∠ C = 90 °, a = 60 °, find the value of sina
∵∠C=90°,∠A=60°
∴∠B=180°-∠C-∠A=30°
Let AC = X
From ab = 2x
∴BC=√(AB²-AC²)=√3x
∴sinA=BC/AB=√3/2
RELATED INFORMATIONS
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