In the triangle ABC, the angle cab120 degrees, ab = 4, AC = 2, ad is perpendicular to BC, and the perpendicular foot is d? Students are in the second year of junior high school, just learning Pythagorean theorem. We hope to solve it in other ways,

This problem has nothing to do with AD;
Make AB vertical line crossing point C and Ba extension line at E;
In the ace of right triangle, if the angle CAE = 60, ACE = 30, then AE = AC / 2 = 1; CE = √ 3;
In BCE of large right triangle, the right side CE = √ 3; be = 4 + 1 = 5;
Then BC 2 = 25 + 3 = 28, that is BC = 2 √ 7;

As shown in the figure, in the right angle △ ABC, ∠ C = 90 ° and the bisector ad of ∠ cab intersects BC at D. if de bisects AB vertically, calculate the degree of ∠ B

∵ in the right angle △ ABC, ∠ C = 90 °, the bisector ad of ∵ cab intersects BC at D,
∴∠DAE=1
2∠CAB=1
2（90°-∠B），
∵ de bisects AB vertically,
∴AD=BD，
∴∠DAE=∠B，
∴∠DAE=1
2∠CAB=1
2（90°-∠B）=∠B，
∴3∠B=90°，
∴∠B=30°．
A: if de bisects AB vertically, the degree of ∠ B is 30 degrees

As shown in the figure, △ ABC, ab = AC, ∠ C = 30 ゜, the vertical bisector Mn of AB intersects BC and ab at points m and N respectively. This paper tries to explore the quantitative relationship between BM and cm

CM=2BM．
Proof: connect am,
In ∵ ABC, ab = AC, ∵ C = 30 ゜,
∴∠B=∠C=30゜，
∵ the vertical bisector Mn of AB intersects BC and ab at points m and N respectively,
∴AM=BM，
∴∠BAM=∠B=30°，
∴∠CAM=180°-∠B-∠C-∠BAM=90°，
∴CM=2AM，
∴CM=2BM．

The triangle ABC angle B = 90 degrees AB = 20 BC = 10 m n are the moving points on AC AB respectively. Find the minimum value of BM + Mn! RT! What is sin? Use what you have learned

Make the symmetry graph AE of AB with respect to AC, make BG perpendicular to G through B, intersect AC with point h. on the straight line AE, BM + Mn = BM + me ≥ be ≥ BG. If and only if M and H coincide, the equal sign holds. Therefore, the minimum value of BM + Mn is = BG = absin ∠ BAE = absin2 ∠ BAC = 2absin ∠ baccos ∠ BAC = 2 * 20 * 10 / 10 √ 5 * 20 / 10 √ 5 = 16

In the triangle ABC, if AB = 3, angle cab = 15 degrees, m and N are the moving points of AC and AB, then the minimum value of BM + Mn is?

Taking AC as the angular bisector, the ray ad is made so that ∠ DAC = ∠ CBA = 15 °;
When B, m and E are on the same line, BM + me is the smallest
∵△ANM≌△AEM,
∴MN=ME.
Then BM + Mn = BM + me = be
Therefore, be ⊥ ad, then △ AEB is a right triangle,
∵∠EAB=2*15°=30°,
∴BE=1/2*AB=1/2*3√2=3√2/2.
The minimum value of BM + Mn is 3 √ 2 / 2

As shown in the figure, in the quadrilateral ABCD, ∠ B = 90 ° ab= 3，∠BAC=30°，CD=2，AD=2 2. Find the degree of ∠ ACD

∵∠B=90°，∠BAC=30°
∴BC=1
2Ac, if BC = x, then AC = 2x
And ∵ ab=
Three
∴（2x）2=x2+（
3）2
∴x=1
∴BC=1，AC=2
CD = 2, ad = 2
Two
∴AC2+CD2=8，AD2=8
∴AC2+CD2=AD2
The △ ACD is a right triangle
∴∠ACD=90°．

As shown in the figure, in the quadrilateral ABCD, ∠ B = 90 ° ab= 3，∠BAC=30°，CD=2，AD=2 2. Find the degree of ∠ ACD

Known, as shown in the figure, in the quadrilateral ABCD, ∠ B = 90 °, ab = BC = radical 6, CD = 3, ad = radical 3, find the degree of: ∠ ACD

In RT △ ABC, ab = BC = root 6, according to Pythagorean theorem, AC = double root 3CD = 3, ad = radical 3, the following four methods: ① according to the inverse theorem of the Pythagorean theorem, there is ∠ ACD = 30 °. ② according to the cosine theorem, there is ∠ ACD = 30 °. ③ sin ∠ ACD = ad / AC = 1 / 2, there is ∠ ACD = 30 °. Cos can also be used

As shown in the figure, in the quadrilateral ABCD, the diagonal lines AC and BD intersect at the point E, BD ⊥ DC, ﹤ abd = 45 °, ACD = 30 ° and ad = CD = 2 3. Find the length of AC and BD

∵BD⊥DC，
∴∠BDC=90°，
∵∠ACD=30°，AD=CD=2
3，
∴∠DEC=60°，∠DAC=∠ACD=30°，
DE=CD•tan30°=2
3 x
Three
3=2，
∴EC=2DE=4，∠ADE=30°，
∴AE=DE=2，
∴AC=AE+EC=2+4=6，
Pass through point a as am ⊥ BD, and perpendicular foot as M,
∵∠AEB=∠DEC=60°，
∴AM=AE•sin60°=2×
Three
2=
3，
ME=AEcos60°=2×1
2=1，
∵∠ABD=45°，
∴BM=AM=
3，
∴BD=BM+ME+DE=
3+1+2=3+
3．

The quadrilateral ABCD is composed of a right triangle ABC with an angle ACB = 30 degrees and an isosceles right triangle ACD. E is the midpoint of AC, and the size of angle BDE is calculated

Connecting be, we can get ∠ ACB = 30 degrees, the distance from the midpoint of the hypotenuse of the right triangle to the three vertices is equal, and AE = be = EC ? the DAC of the triangle is an isosceles right triangle ? DC = Da, ∠ Dec = 90 degrees, ﹤ DAE = 45 degrees  EDA = 45 degrees  AED is an isosceles right triangle ? AE = be = de  triangle

In the triangle ABC, the angle cab120 degrees, ab = 4, AC = 2, ad is perpendicular to BC, and the perpendicular foot is d? Students are in the second year of junior high school, just learning Pythagorean theorem. We hope to solve it in other ways,