In triangle abc ,if ab=ac and ab is produced to d such that bd=bc, find angle acd:angle adc
Answers
» ΔABC in which AB is extended to D such that
=> BD = BC
Also
=> AB = AC
_____________
=> ∠ABC = ∠ACB [Angles opposite to equal sides are equal]
» Again,
=> BD = BC
=> ∠BDC = ∠BCD [Angles opposite to equal sides are equal]
_______________
» Now,
» in ΔABC by angle sum property of triangle :-
=> ∠BAC + ∠ABC + ∠ACB = 180º
=> ∠BAC = 180º - (∠ACB + ∠ABC) = 180º - (2∠ACB)
=> ∠BAC = 180º - (2∠ACB) ___ [Eq1]
__________________
» Again in ΔADC by angle sum property of triangle :-
=> ∠DAC +∠ADC + ∠ACD = 180º
=> 180º - (2∠ACB) +∠ADC + ∠ACD = 180º [Using (1), as ∠BAC = ∠DAC = 180º - (2∠ACB)]
=> - 2∠ACB +∠ADC + ∠ACB + ∠BCD = 0
=> - ∠ACB +∠ADC + ∠BCD = 0
=> - ∠ACB +∠ADC + ∠ADC = 0 [As, ∠ADC =∠BDC = ∠BCD]
=> - ∠ACB + 2∠ADC = 0
=> ∠ACB = 2∠ADC ____[Eq2]
__________________
» Now,
=> ∠ACD/∠ADC = (∠ACB+∠BCD)/∠ADC
» By Eq2 ,
=> ∠ACD/∠ADC = (2∠ADC+∠ADC)/∠ADC
=> ∠ACD/∠ADC = 3∠ADC/∠ADC
=> ∠ACD/∠ADC = 3/1
» THEN,
=> ∠ACD : ∠ADC = 3:1
__________________[ANSWER]
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_-_-_-_✌☆
●Given,
» ΔABC in which AB is extended to D such that
=> BD = BC
Also
=> AB = AC
_____________
=> ∠ABC = ∠ACB [Angles opposite to equal sides are equal]
» Again,
=> BD = BC
=> ∠BDC = ∠BCD [Angles opposite to equal sides are equal]
_______________
» Now,
» in ΔABC by angle sum property of triangle :-
=> ∠BAC + ∠ABC + ∠ACB = 180º
=> ∠BAC = 180º - (∠ACB + ∠ABC) = 180º - (2∠ACB)
=> ∠BAC = 180º - (2∠ACB) ___ [Eq1]
__________________
» Again in ΔADC by angle sum property of triangle :-
=> ∠DAC +∠ADC + ∠ACD = 180º
=> 180º - (2∠ACB) +∠ADC + ∠ACD = 180º [Using (1), as ∠BAC = ∠DAC = 180º - (2∠ACB)]
=> - 2∠ACB +∠ADC + ∠ACB + ∠BCD = 0
=> - ∠ACB +∠ADC + ∠BCD = 0
=> - ∠ACB +∠ADC + ∠ADC = 0 [As, ∠ADC =∠BDC = ∠BCD]
=> - ∠ACB + 2∠ADC = 0
=> ∠ACB = 2∠ADC ____[Eq2]
__________________
» Now,
=> ∠ACD/∠ADC = (∠ACB+∠BCD)/∠ADC
» By Eq2 ,
=> ∠ACD/∠ADC = (2∠ADC+∠ADC)/∠ADC
=> ∠ACD/∠ADC = 3∠ADC/∠ADC
=> ∠ACD/∠ADC = 3/1
» THEN,
=> ∠ACD : ∠ADC = 3:1