Square defg is the inscribed square of triangle ABC. Am is perpendicular to BC, m and intersects DG at h. If ah is 4cm long and the side of the square is 6cm long, find B Square defg is the inscribed square of triangle ABC. Am is perpendicular to BC, m and intersects DG at h. If ah is 4cm long and the side length of the square is 6cm, the length of BC is calculated

Square defg is the inscribed square of triangle ABC. Am is perpendicular to BC, m and intersects DG at h. If ah is 4cm long and the side of the square is 6cm long, find B Square defg is the inscribed square of triangle ABC. Am is perpendicular to BC, m and intersects DG at h. If ah is 4cm long and the side length of the square is 6cm, the length of BC is calculated

The length of BC is 15
Do it in a similar way

As shown in the figure, the square defg is the inscribed square of △ ABC, am ⊥ BC in M, intersecting DG in H. if ah is 4cm long, the side length of the square is 6cm. Find the length of BC

If ah is 4cm long and the side of a square is 6cm long, find the length of BC www.dg/ , recently discovered

In RT △ ABC, ∠ C = 90 °, a right triangle is solved by the following conditions
(1) Given a = 20, C = 20 √ 2, find ∠ B
(2) Given C = 30, ∠ a = 60 °, find a
Evaluation: Sin & # 178; 30 ° + cos & # 178; 30 ° + Tan & # 178; 60 ° cot & # 178; 60 °

1. In RT △ ABC, ∠ C = 90 ° C = 20 √ 2, a = 20
cosB=a/c=√2 /2
∠ B=45 °
2. In RT △ ABC, ∠ C = 90 °, a = 60 °
sin60° =a/c=√3 /2
a=15=√3
3、sin30°=cos60°=1/2
cos30°=sin60°=√3 /2
tan60 °=sin60°/cos60°=√3
cot60°=1/tan60°=√3 /3
So Sin & # 178; 30 ° + cos & # 178; 30 ° + Tan & # 178; 60 ° cot & # 178; 60 ° = 1 / 4 + 3 / 4 + 9 = 10

In ABC AB = AC de parallel BC angle EDF = Abe prove DG times DF = DB times ef
In ABC AB = AC de parallel BC angle EDF = Abe prove DG times DF = DB times ef

AB=AC,DE//BC
So ad = AE, ∠ ade = ∠ AED,
So, BDE = fed,
And ∠ EDF = ∠ Abe
Therefore, EFD = bed,
To sum up, EDF = Abe, BDE = fed, EFD = bed
So △ def ∽ BDE
∠ABE=∠EDF,∠DEG=∠DEB,∠DGE=∠BDE
So △ BDE ∽ DGE
So dg / BD = de / be,
And because △ def ∽ BDE (proved)
So EF / DF = de / be
So dg / BD = EF / DF
So DG * DF = BD * ef