Giả sử cạnh của tam giác đều ABCcap A cap B cap C
𝐴𝐵𝐶
là 2a2 a
2𝑎
. Khi đó, đường cao AH=a3cap A cap H equals a the square root of 3 end-root
𝐴𝐻=𝑎3√
và HC=acap H cap C equals a
𝐻𝐶=𝑎
.
Theo giả thiết, DH=AH=a3cap D cap H equals cap A cap H equals a the square root of 3 end-root
𝐷𝐻=𝐴𝐻=𝑎3√
.
Xét điểm E′cap E prime
𝐸′
trên tia ACcap A cap C
𝐴𝐶
sao cho CE′=HC=acap C cap E prime equals cap H cap C equals a
𝐶𝐸′=𝐻𝐶=𝑎
. Ta sẽ chứng minh E′cap E prime
𝐸′
trùng với Ecap E
𝐸
. 
Trong ΔHCE′cap delta cap H cap C cap E prime
Δ𝐻𝐶𝐸′
, ta có HC=CE′=acap H cap C equals cap C cap E prime equals a
𝐻𝐶=𝐶𝐸′=𝑎
và ∠HCE′=180∘−∠ACB=120∘angle cap H cap C cap E prime equals 180 raised to the composed with power minus angle cap A cap C cap B equals 120 raised to the composed with power
∠𝐻𝐶𝐸′=180∘−∠𝐴𝐶𝐵=120∘
(do A,C,E′cap A comma cap C comma cap E prime
𝐴,𝐶,𝐸′
thẳng hàng).
ΔHCE′cap delta cap H cap C cap E prime
Δ𝐻𝐶𝐸′
cân tại C⇒∠CHE′=∠CE′H=180∘−120∘2=30∘cap C implies angle cap C cap H cap E prime equals angle cap C cap E prime cap H equals the fraction with numerator 180 raised to the composed with power minus 120 raised to the composed with power and denominator 2 end-fraction equals 30 raised to the composed with power
𝐶⇒∠𝐶𝐻𝐸′=∠𝐶𝐸′𝐻=180∘−120∘2=30∘
.
Áp dụng định lý hàm số cos trong ΔHCE′cap delta cap H cap C cap E prime
Δ𝐻𝐶𝐸′
:
HE′2=HC2+CE′2−2HC⋅CE′⋅cos(120∘)=a2+a2−2a2(−12)=3a2cap H cap E prime squared equals cap H cap C squared plus cap C cap E prime squared minus 2 cap H cap C center dot cap C cap E prime center dot cosine open paren 120 raised to the composed with power close paren equals a squared plus a squared minus 2 a squared open paren negative one-half close paren equals 3 a squared
𝐻𝐸′2=𝐻𝐶2+𝐶𝐸′2−2𝐻𝐶⋅𝐶𝐸′⋅cos(120∘)=𝑎2+𝑎2−2𝑎2(−12)=3𝑎2

Suy ra HE′=a3cap H cap E prime equals a the square root of 3 end-root
𝐻𝐸′=𝑎3√
. Vì DH=a3cap D cap H equals a the square root of 3 end-root
𝐷𝐻=𝑎3√
nên HE′=DHcap H cap E prime equals cap D cap H
𝐻𝐸′=𝐷𝐻
.
Xét ΔDHE′cap delta cap D cap H cap E prime
Δ𝐷𝐻𝐸′
, ta có HD=HE′=a3cap H cap D equals cap H cap E prime equals a the square root of 3 end-root
𝐻𝐷=𝐻𝐸′=𝑎3√
nên ΔDHE′cap delta cap D cap H cap E prime
Δ𝐷𝐻𝐸′
cân tại Hcap H
𝐻
.
Góc ∠DHE′=180∘−∠CHE′=180∘−30∘=150∘angle cap D cap H cap E prime equals 180 raised to the composed with power minus angle cap C cap H cap E prime equals 180 raised to the composed with power minus 30 raised to the composed with power equals 150 raised to the composed with power
∠𝐷𝐻𝐸′=180∘−∠𝐶𝐻𝐸′=180∘−30∘=150∘
.
⇒∠HDE′=180∘−150∘2=15∘implies angle cap H cap D cap E prime equals the fraction with numerator 180 raised to the composed with power minus 150 raised to the composed with power and denominator 2 end-fraction equals 15 raised to the composed with power
⇒∠𝐻𝐷𝐸′=180∘−150∘2=15∘
.
Vì ∠HDE′=15∘angle cap H cap D cap E prime equals 15 raised to the composed with power
∠𝐻𝐷𝐸′=15∘
trùng với giả thiết ∠BDx=15∘angle cap B cap D x equals 15 raised to the composed with power
∠𝐵𝐷𝑥=15∘
và Ecap E
𝐸
thuộc tia ACcap A cap C
𝐴𝐶
nên E≡E′cap E triple bar cap E prime
𝐸≡𝐸′
.
Vậy EH=DHcap E cap H equals cap D cap H
𝐸𝐻=𝐷𝐻
và HC=ECcap H cap C equals cap E cap C
𝐻𝐶=𝐸𝐶
. 

Bước 2: Chứng minh EH⟂ABcap E cap H ⟂ cap A cap B
𝐸𝐻⟂𝐴𝐵
 
Gọi Kcap K
𝐾
là giao điểm của đường thẳng EHcap E cap H
𝐸𝐻
và ABcap A cap B
𝐴𝐵
. 
Từ kết quả câu (a), ta có ∠CHE=30∘angle cap C cap H cap E equals 30 raised to the composed with power
∠𝐶𝐻𝐸=30∘
.
Vì Hcap H
𝐻
nằm giữa Bcap B
𝐵
và Ccap C
𝐶
, tia HEcap H cap E
𝐻𝐸
nằm ở nửa mặt phẳng dưới BCcap B cap C
𝐵𝐶
, nên góc đối đỉnh của ∠CHEangle cap C cap H cap E
∠𝐶𝐻𝐸
là góc tạo bởi tia đối của HEcap H cap E
𝐻𝐸
và tia HBcap H cap B
𝐻𝐵
. Gọi góc này là ∠BHK=30∘angle cap B cap H cap K equals 30 raised to the composed with power
∠𝐵𝐻𝐾=30∘
.
Trong ΔABCcap delta cap A cap B cap C
Δ𝐴𝐵𝐶
đều, ∠ABC=60∘angle cap A cap B cap C equals 60 raised to the composed with power
∠𝐴𝐵𝐶=60∘
hay ∠KBH=60∘angle cap K cap B cap H equals 60 raised to the composed with power
∠𝐾𝐵𝐻=60∘
.
Xét tam giác BKHcap B cap K cap H
𝐵𝐾𝐻
, ta có:
∠BKH=180∘−(∠KBH+∠BHK)=180∘−(60∘+30∘)=90∘angle cap B cap K cap H equals 180 raised to the composed with power minus open paren angle cap K cap B cap H plus angle cap B cap H cap K close paren equals 180 raised to the composed with power minus open paren 60 raised to the composed with power plus 30 raised to the composed with power close paren equals 90 raised to the composed with power
∠𝐵𝐾𝐻=180∘−(∠𝐾𝐵𝐻+∠𝐵𝐻𝐾)=180∘−(60∘+30∘)=90∘

Vậy EH⟂ABcap E cap H ⟂ cap A cap B
𝐸𝐻⟂𝐴𝐵
tại Kcap K
𝐾
. 

Đáp án: 
a. EH=DHcap E cap H equals cap D cap H
𝐸𝐻=𝐷𝐻
(cùng bằng AHcap A cap H
𝐴𝐻
)
b. HC=ECcap H cap C equals cap E cap C
𝐻𝐶=𝐸𝐶

c. EH⟂ABcap E cap H ⟂ cap A cap B
𝐸𝐻⟂𝐴𝐵
tại giao điểm của chúng. 

 
 
 
 
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