∠A = 180 -(∠B +∠C) or
∠A = 180 - 3*(∠C)-------------------------------- 1.
But its should be given that AD = DC (and not AB = BC as given) So as AD is bisector of A
(∠A )/2 = ∠C ---------------------------- 2 or multiplying by 3 we get
(3/2)∠A = 3∠C -------------------------- 3
Adding 1 and 3, we get
(5/2)∠A = 180 or
∠A or ∠BAC = 360/5 = 72
hence proved.
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∠A = 180 -(∠B +∠C) or
∠A = 180 - 3*(∠C)-------------------------------- 1.
But its should be given that AD = DC (and not AB = BC as given) So as AD is bisector of A
(∠A )/2 = ∠C ---------------------------- 2 or multiplying by 3 we get
(3/2)∠A = 3∠C -------------------------- 3
Adding 1 and 3, we get
(5/2)∠A = 180 or
∠A or ∠BAC = 360/5 = 72
hence proved.