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In 3-angle ABC, the angle BAC is 90 degrees, ad is perpendicular to BC, EF is perpendicular to BC, FM is perpendicular to AC, the perpendicular foot is D, F, m respectively, angle 1 = angle 2, verification: FM = FD
Give me a picture
How to answer without a picture? This question is very simple
But where's corner 1 and corner 2? Give me a picture
Given △ ABC, ∠ BAC = 90 °, ad ⊥ BC in D, EF ⊥ BC, FM ⊥ AC in M, ∠ 1 = ∠ 2, FM = FD is proved
Can you give me a picture
As shown in the figure, it is known that in △ ABC, ∠ BAC + 90 °, ad ⊥ BC, EF ⊥ BC, FM ⊥ AC, ∠ 1 = ∠ 2, FM = FD
And the picture I can't help you
It is known that in the triangle ABC, ∠ a = 90 °, ab = AC, and D is the midpoint of BC. (1) as shown in the figure, e and F are the points on AB and AC respectively, and be = AF, and it is proved that △ DEF is an isosceles right triangle; (2) if e and F are the points on the extension line of AB and Ca respectively, and be = AF still exists, and other conditions remain unchanged, is △ def still an isosceles right triangle? Prove your conclusion
(1) It is proved that: connecting ad, ∵ AB = AC, ≌ BAC = 90 °, D is the midpoint of BC, ∵ ad ⊥ BC, BD = ad. ∵ B = ≌ DAC = 45 ° and be = AF, ≌ BDE ≌ ADF (SAS) ≌ ed = FD, ≌ BDE = ≌ ADF. ≁ EDF = ≁ EDA + ≁ ADF = ≁ EDA + ≁ ADF = ≌ BDA = 90 °. DEF is isosceles right triangle. (2) ≌ DEF is isosceles right triangle. It is proved that if e and F are AB and Ca respectively The points on the line are as shown in the figure: connect ad, ∫ AB = AC, ∫ ABC is isosceles triangle, ∫ BAC = 90 °, D is the midpoint of BC, ∫ ad = BD, ad ⊥ BC (three lines in one), ∫ DAC = ≁ abd = 45 °, and ∫ DAF = ≌ DBE = 135 °. AF = be, ≌ DAF ≌ DBE (SAS).. FD = ed, ≁ FDA = ≁ EDB. ≁ EDF = ≁ EDB + ≁ FDB = ≁ FDA + ≁ FDB = ≁ ADB = 90 °. DEF is still isosceles straight Angle triangle
In △ ABC, AB & # 178; = 90, BC & # 178; = 9, AC & # 178;, then AC & # 178; = ()
It's a right triangle, right
In △ ABC, ∠ C = 90 °, AB & # 178; = 18, and AC = BC, then BC=____ .
We need to be specific
It is known that ∠ C = 90 ° so AB is a hypotenuse
According to Pythagorean theorem, AC & # 178; + BC & # 178; = AB & # 178;
Known AC = BC
∴2BC²=18
∴BC=3
In the isosceles right triangle ABC, the bisector of ∠ a = 90 & # 186; ∠ BAC intersects BC at EEF ⊥ AC at FFG ⊥ AB at g. it is proved that ab & # 178; = 2fg & # 178;
"FG ⊥ AB at g" is changed to "the extension line of FG ⊥ BC intersects BA at point G"
Because AB = AC, AE bisects ∠ BAC, so AE ⊥ BC, and be = CE,
Because of FG ⊥ BC, AE ∥ FG;
Because ∠ a = 90 & # 186;, EF ⊥ AC, Ag ∥ EF,
So, the quadrilateral aefg is a parallelogram, so FG = AE
Because AB = AC, ∠ a = 90 & # 186;, BC & # 178; = AB & # 178; + AC & # 178; = 2Ab & # 178
Because ∠ a = 90 & # 186;, be = CE, so BC = 2ae,
So, (2ae) &# 178; = 2Ab & # 178;,
AB²=2AE²=2FG².
In △ ABC, ab = M & # 178; + 1, BC = 2m, AC = M & # 178; - 1, where m is an integer greater than 1, find the size of angle ACB//
∵AC^2＋BC^2＝（m^2－1）^2＋（2m）^2＝（m^2＋1）^2, AB^2＝（m^2＋1）^2
∴AC^2＋BC^2＝AB^2.
From the converse theorem of Pythagorean theorem, it is obtained that: ∠ ACB = 90 °
Acute angle trigonometric function: (1) in RT Δ ABC, ∠ C = 90 °, BC = 2Mn, AC = M & # 178; - N & # 178; (M > n > 0), find the three trigonometric function values of ∠ B
Acute trigonometric function:
(1) In RT Δ ABC, ∠ C = 90 °, BC = 2Mn, AC = M & # 178; - N & # 178; (M > n > 0), three trigonometric functions of ∠ B are obtained
(2) It is known that √ 2 + 1 is a root of the equation x & # 178; - 3tan & # 216; · x + √ 2 = 0, &# 216; is an internal angle of the triangle, and the value of COS & # 216; is obtained
(1) AB=√(AC²+BC²)
=√[(m²-n²)²+(2mn)²]
=m²+n²
∴sinB=AC/AB=(m²-n²)/(m²+n²)
cosB=BC/AB=2mn/(m²+n²)
tanB=AC/BC=(m²-n²)/2mn
(2) Let the other root be x1
∴(√2+1)x1=√2
∴x1=2-√2
x1+(√2+1)=3tanØ
∴tanØ=1
∴Ø=45°
∴cosØ=cos45°=√2/2
In the triangle ABC, CD is the height on the edge of AB, if a & # 178; + C & # 178; < B & # 178;, CD & # 178; / AC & # 178; + CD & # 178; / BC & # 178; = 1
Then we prove that B-A = π / 2
I'm glad to answer your question CD & # AC & # 178; + CD & # 178; + BC & # 178; = 1sin & # 178; a + Sin & # 178; b = 1sin & # 178; b = 1-sin & # 178; a = cos & # 178; a, SINB = cosaa & # 178; + C & # 178; < B & # 178;, a & lt; B, a & lt; bb-a = π / 2
RELATED INFORMATIONS
- 1. As shown in the figure, it is known that CD is perpendicular to ab at point D, be is perpendicular to AC at point E, and CD intersects be at point o (1) If OC = ob, prove that point O is on the bisector of angular BAC (2) If point O is on the bisector of angle BAC, OC = ob is proved
- 2. In the triangle ABC, ab = AC, D is a point on BC, and
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- 20. The three sides of triangle are equal difference series, the circumference is 36, and the circumference of inscribed circle is 6 π Some people do that ∵ the three sides grow into an arithmetic sequence with a circumference of 36 Let the three sides be 12-n, 12, 12 + n respectively The circumference of the inscribed circle is 6 π --- radius of the inscribed circle r = 3 --- triangle area = (36 / 2) * 3 = 54 From Helen's formula: 54 & sup2; = 18 * 6 * (6 + n) * (6-n) --->27=36-n²--->n²=9--->n=3 --->The lengths of the three sides are 9, 12 and 15 respectively ∵ 9 & sup2; + 12 & sup2; = 15 & sup2;, ∵ the triangle is a right triangle Then why must one side be 12?