Question

ABC is an isosceles triangle such that AB = AC and AD is the median to base BC. if <B =35° then find angle BAD.​

Answer 1

Question:-

• ABC is an isosceles triangle such that AB = AC and AD is the median to base BC. if ∠B =35° then find angle BAD.

Given:

• ABC, is an isosceles triangle such that AB = AC and AD is the median to base BC.

 

To Find :

• ∠BAD

Answer:-

• 55°

 

Proof :

• From figure we have ∠ABC = 35°  
• ∆ABC an isosceles triangle and  AB = AC  

∠ABC = ∠ACB = 35°

As we know that ,

$\sf \: Angles \: opposite \: to \: equal \: sides \: of \: a \: triangle \: are \: equal.$

• Let ∠ADB be x.

∠ADB + ∠ADC = 180° [Linear pair]

x + ∠ADC = 180°

$\sf ( ∠ADC = 180° - x ) - - - - - - - - - (1)$

 

• 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 =
• in isosceles triangle AB = AC.

$\frac{AB}{AC} = \frac{BD}{CD} = 1$

$\bf \: Let \: ∠BAD = ∠CAD = y$

In ∆BAD,  

Since Sum of the angles of a triangle is 180° :  

∠ABD + ∠ADB + ∠BAD = 180°

$\bf \: (35° + x + y = 180°) - - - - - - (2)$

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°

$\bf \: ⇝y = \frac{100°}{2}$

⇝y = 55°

⇒∠BAD =  55°  

Hence, ∠BAD is  55° .