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The sides BA and BC of ΔABC are produced to D and E respectively. Prove that ∠DAC + ∠ECA > 180°. - Mathematics

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प्रश्न

The sides BA and BC of ΔABC are produced to D and E respectively. Prove that ∠DAC + ∠ECA > 180°.

प्रमेय
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उत्तर

Given: The sides BA and BC of ΔABC are produced to points D and E respectively.

To Prove: ∠DAC + ∠ECA > 180°

Proof:

The sides BA and BC of ΔABC are produced to D and E respectively.

∠DAC = ∠B + ∠ACB    ...(By exterior angle property)

∠ECA = ∠B + ∠BAC    ...(By exterior angle property)

And ∠ECA + ∠ACB = 180°   ...(Linear pairs)

∠ECA = 180° – ∠ACB   ...(1)

∠DAC + ∠BAC = 180°   ...(Linear pairs)

∠DAC = 180° – ∠BAC   ...(2)

Adding (1) and (2),

∠ECA + ∠DAC = 180° – ∠ACB + 180° – ∠BAC

∠ECA + ∠DAC = 360° – ∠ACB – ∠BAC

∠ECA + ∠DAC = 360° – (∠ACB + ∠BAC)

∠ECA + ∠DAC = 360° – (180° – ∠ABC)   ...(Angle sum property of triangle)

∠ECA + ∠DAC = 360° – 180° + ∠ABC 

∠ECA + ∠DAC = 180° + ∠ABC

So, ∠ECA + ∠DAC > 180°

Hence, proved.

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अध्याय 9: Inequalities - EXERCISE 9 [पृष्ठ १०३]

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बी निर्मला शास्त्री Mathematics [English] Class 9 ICSE
अध्याय 9 Inequalities
EXERCISE 9 | Q 14. | पृष्ठ १०३
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