Để chứng minh tam giác ABC đồng dạng với tam giác ACD, ta cần chứng minh hai tam giác này có một cặp góc bằng nhau và có một cặp cạnh tương ứng tỉ lệ với nhau.
Ta có AB = 9cm, AC = 12cm, BC = 7cm và BD = BC = 7cm.
Do BD = BC nên tam giác BCD là tam giác cân tại B.
Do đó, góc BCD = góc BDC.
Ta cũng có góc ABC = góc ACD vì chúng là góc ở tia đối của tia BA.
Vậy ta có hai cặp góc bằng nhau giữa hai tam giác ABC và ACD.
Để tính độ dài đoạn thẳng CD, ta sử dụng định lý cosin trong tam giác ACD:
AC^2 = AD^2 + CD^2 - 2*AD*CD*cos(ACD)
12^2 = AD^2 + CD^2 - 2*AD*CD*cos(ACD)
144 = AD^2 + CD^2 - 2*AD*CD*cos(ACD) (1)
Do tam giác ABC đồng dạng với tam giác ACD nên ta có tỉ lệ cạnh:
AB/AC = BC/CD
9/12 = 7/CD
CD = 12*7/9 = 84/9 = 28/3 cm
Thay CD = 28/3 vào (1) ta có:
144 = AD^2 + (28/3)^2 - 2*AD*(28/3)*cos(ACD)
144 = AD^2 + 784/9 - 56/3*cos(ACD)
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144 = AD^2 + 784/9 - 56/3*cos(ACD)
144 = AD^2 +