Question

solve please let me know​

Answer 1

Given :

• AB = BD
• DC = AD
• Angle ACD = 40°

To find :

Angle ABC

Solution :

In ∆ADC , AD = DC

therefore ∆ADC is isosceles triangle with

➝ angle(DAC) = angle(ACD)

➝ angle(DAC) = 40°

• Also, angle(ADB) is exterior angle of ∆ADC,

{ By property➝ exterior angle = Sum of opposite interior angle }

➝ angle(ADB) = angle(DAC) + angle(ACD)

➝ angle(ADB) = 40° + 40°

➝ angle(ADB) = 80°

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In ∆ABD , AB = BD

therefore, ∆ABD is isosceles triangle,with

➝ angle(ADB) = angle(BAD)

➝ angle(BAD) = 80°

Now , we know that sum of interior angle of triangle = 180°

➝ angle(ADB) + angle(BAD) + angle(ABD) = 180°

➝ 80° + 80° + angle(ABD) = 180°

➝ angle(ABD) = 180° - 160°

➝ angle(ABD) = 20°

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As , shown in image BC is extension of line BD to a point C

therefore,

angle(ABC) = angle(ABD)

angle(ABC) = 20°

ANSWER :

angle(ABC) = 20°

Answer 2

The answer is 20°.

Hope this helps