Q16 of 45 Page 164

In the adjoining figure,

Prove that


∠ABC – ∠DCB + ∠CDE = 180°.


3.JPG


⇒ ∠ABC =180° - ∠GBC =180° - ∠DHC …… (1)


And


⇒ ∠CDE = 180°- ∠HDC …… (2)


From (1) + (2) we get,


⇒ ∠ABC + ∠CDE = 180° - ∠DHC + 180°- ∠HDC


⇒ ∠ABC + ∠CDE = 360° - (∠DHC + ∠HDC)


⇒ ∠ABC + ∠CDE = 360° - (180° - ∠HCD) [from ΔDHC]


⇒ ∠ABC + ∠CDE = 180° + ∠DCB


⇒ ∠ABC – ∠DCB + ∠CDE = 180°


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