问一道初一的数学题!~急啊

连接BC,AD
则根据三角形内角和为180度,所以
∠BDC + ∠DBC + ∠DCB = 180
因此 ∠BDC = 180 - ∠DBC - ∠DCB
= 180 - (∠ABC - ∠ABD) - (∠ACB - ∠ACD)
= (180 - ∠ABC - ∠ACB) + ∠ABD + ∠ACD
= ∠A + ∠ABD + ∠ACD
同上
∠BDC = ∠BGC + ∠GBD + ∠GCD
= ∠BGC + 1/2 (∠ABD + ∠ACD)
故∠ABD + ∠ACD = ∠BDC - ∠A = 2 (∠BDC - ∠BGC)
即 ∠BDC + ∠A = 2∠BGC
这就是所求的三个角之间的关系和证明