Question

in a triangle ABC AB=AC is produced to D. sich that BD=BC. prove that Angle ACD=thrice of angle ADC​

Answer 1

Answer:

in the triangle

Step-by-step explanation:

in the triangle ABC AB=AC is a produced to D

Answer 2

●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