ABC, is an isosceles triangle such that AB=AC and AD is the median to base BC. Then, ∠BAD=
A. 55°
B. 70°
C. 35°
D. 110°
Answers
Given: ABC, is an isosceles triangle such that AB = AC and AD is the median to base BC.
To Find : ∠BAD
Proof :
From figure we have ∠ABC = 35°
we have , ∆ABC an isosceles triangle and
AB = AC
∠ABC = ∠ACB = 35°
[ Angles opposite to equal sides of a triangle are equal]
Let ∠ADB be x.
∠ADB + ∠ADC = 180° [Linear pair]
x + ∠ADC = 180°
∠ADC = 180° - x ..............(1)
By using angle bisector theorem,if a ray bisects an angle of a triangle, then it divides the opposite side into segments that are proportional to the other two sides :
AD is median so BD = CD and in isosceles triangle AB = AC.
AB/AC = BD/CD = 1
Let ∠BAD = ∠CAD = y
In ∆BAD,
Since Sum of the angles of a triangle is 180° :
∠ABD + ∠ADB + ∠BAD = 180°
35° + x + y = 180° ………..(2)
In ∆DAC,
Since Sum of the angles of a triangle is 180° :
∠ACD + ∠ADC + ∠CAD = 180°
35° + 180° - x + y = 180°
[From eq 1]
35° - x + y = 180° - 180°
35 + y = x
Put this value of x in eq 2,
35° + x + y =180°
35° + 35 + y + y =180°
2y + 70 = 180°
2y = 180° - 70°
2y = 100°
y = 100°/2
y = 55°
∠BAD = 55°
Hence, ∠BAD is 55° .
Among the given options option (A) 55° is correct.
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Answer:
Step-by-step explanation:
we have ∠ABC = 35°
we have , ∆ABC an isosceles triangle and
AB = AC
∠ABC = ∠ACB = 35°
[ Angles opposite to equal sides of a triangle are equal]
Let ∠ADB be x.
∠ADB + ∠ADC = 180° [Linear pair]
x + ∠ADC = 180°
∠ADC = 180° - x ..............(1)
AD is median so BD = CD and in isosceles triangle AB = AC.
AB/AC = BD/CD = 1
Let ∠BAD = ∠CAD = y
In ∆BAD,
Since Sum of the angles of a triangle is 180° :
∠ABD + ∠ADB + ∠BAD = 180°
35° + x + y = 180° ………..(2)
In ∆DAC,
Since Sum of the angles of a triangle is 180° :
∠ACD + ∠ADC + ∠CAD = 180°
35° + 180° - x + y = 180°
[From eq 1]
35° - x + y = 180° - 180°
35 + y = x
Put this value of x in eq 2,
35° + x + y =180°
35° + 35 + y + y =180°
2y + 70 = 180°
2y = 180° - 70°
2y = 100°
y = 100°/2
y = 55°
∠BAD = 55°