Question

超难数学题！！！！！！！！！！！！！

Answer 1

两角相等
∠CIH=90-(1/2)∠C
∠BID=180-∠IDB-(1/2)∠B
∠IDB=180-∠B-(1/2)∠A
故∠BID=180-(180-∠B-(1/2)∠A)-(1/2)∠B
=∠B+(1/2)∠A-(1/2)∠B
=(1/2)∠B+(1/2)∠A
=(1/2)(180-∠C)
=90-(1/2)∠C
∠A+∠B+∠C=180
有不懂欢迎追问

Answer 2

<BID=1/2(<A+<B)=1/2(180-<C)=90-1/2<C
<CIH=90-1/2<C (在直角三角形ICH中可得）
所以两个角相等

Answer 3

两角相等
∠CIH=90-(1/2)∠C
∠BID=180-∠IDB-(1/2)∠B
∠IDB=180-∠B-(1/2)∠A
故∠BID=180-(180-∠B-(1/2)∠A)-(1/2)∠B
=∠B+(1/2)∠A-(1/2)∠B
=(1/2)∠B+(1/2)∠A
=(1/2)(180-∠C)
=90-(1/2)∠C
∠A+∠B+∠C=180