AD is the median of an isosceles triangle ABC in which AB = AC and ∠B = 40 then find ∠ADC.
Answers
In ΔABC
AB = AC so ∠B must be equal to ∠C.......(∠B=∠C=40°)
∠A=180°-∠B-∠C..................................(angle sum property)
∠A=180°-40°-40°
∠A=100°
Median of a triangle bisects the vertex angle so
∠DAC+∠DAB=100°
⇒2(∠DAC)=100°
⇒∠DAC=50°
IN ΔACD:-
∠A+∠C+∠D=180°.......(ANGLE SUM PROPERTY)
50°+40°+∠D=180°
90°+∠D=180°
∠D=90°
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Given : ∆ABC is an isosceles triangle such that AB = AC and
AD is median to BC.
∠B = 40°
To Find : ∠adc
Solution:
∆ BAD and ∆ CAD
AB = AC Given
AD = AD Common
BD = CD = BC/2 as AD is the median
Using side - side - side congruence property
∆ BAD ≅ ∆ CAD
=> ∠ADC ≅ ∠ADB
AB is straight line
Hence ∠ADC and ∠ADB form a linear pair
=> ∠ADC + ∠ADB = 180°
=> 2∠ADC = 180°
=>∠ADC = 90°
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