Question

Suppose ABC is an equilateral triangle it base BC as produced to D. Such that BC = CD. Calculate \angle ACD, \Angle ADc​

Answer 1

Answer:

$\bold\underline\huge\underline{\pink{\boxed{\mathfrak{Given}}}}$

• ${\triangle}$ ABC is an equilateral triangle
• BC = CD

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$\bold\underline\huge\underline{\pink{\boxed{\mathfrak</p><p>{To \: prove}}}}$

• (1)${\angle}$ACD and (2)${\angle}$ADC

• ${\green}$Construction : Join CD & AD

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In ${\triangle}$ABC

${\angle}$ A + ${\angle}$ B + ${\angle}$ C =180 ${\degree}$

x + x + x = 180${\degree}$

3x =180 ${\degree}$

x = ${\dfrac{180 \degree}{3}}$

x = 60${\degree}$

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• ${\angle}{\green}$A = ${\angle}{\green}$B = ${\angle}{\green}$C = 60${\degree}{\green}$

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• BC = AC – [1]
• BC = CD – [2]

• AC = CD [from equation 1]

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Interior(C) + Exterior(C) = 180${\degree}$

60${\degree}$ + x = 180${\degree}$

x = 180${\degree}$ - 60${\degree}$= 120 ${\degree}$

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In ${\triangle}$ ACD

${\angle}$ A + ${\angle}$C + ${\angle}$ D = 180 ${\degree}$

y + 120${\degree}$ + y = 180 ${\degree}$

2y + 120 ${\degree}$ = 120 ${\degree}$ = 180

2y = 180 ${\degree}$ - 120 ${\degree}$ = 60 ${\degree}$

y = ${\dfrac{60 \degree}{2}}$ =30${\degree}$

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• ${\angle}{\color}{\green}$ A = y = 30${\degree}$
• ${\angle}{\green}$ D = y = 30${\degree}$