In the triangle ABC, ab = AC, CD is the height of edge AB, and it is proved that ∠ BCD = half ∠ a
The proof is as follows:
∠A=180-2∠ACB
∠A=2(90-∠ACB)
∠A=2{90-(∠DCB+∠ACD)}
And ∠ ACD = 90 - ∠ a
Bring in the above formula
∠A=2(∠A-∠DCB)
It is proved that: 2 ∠ DCB = ∠ a
The above 180 and 90 are all degrees
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