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Question 41

In ΔABC, A=50,B=70 and bisector of C meets AB at D as shown in the given figure. Measure of ADC is

a) 50
b) 100
c) 30
d) 70


Solution

In ΔADC,
ADC+DAC+ACD=180    [angle sum property of a triangle]
ADC+50+ACD=180    [DAC=50]
ACD=130ADC(i)
In ΔDBC,ADC=DBC+BCD
[ exterior angle is equal to sum of opposite interior angles]
ADC=70+ACD    [ACD=BCD]
ADC=70+130ADC   [from Eq. (i)]
ADC=200ADC2ADC=200ADC=2002ADC=100

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