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pratu043

  • 3 years ago

In a triangle ABC, AD is the bisector of angle BAC and angle C is greater than angle B. Prove that angle ADB > angle ADC.

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  1. pratu043
    • 3 years ago
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    |dw:1348233132884:dw|

  2. pratu043
    • 3 years ago
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    Please help I have an exam tomorrow and need to get my doubts cleared.

  3. pratu043
    • 3 years ago
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    @lgbasallote please please help ....

  4. pratu043
    • 3 years ago
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    @lgbasallote help!!!!!

  5. pratu043
    • 3 years ago
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    Can you help @Mimi_x3?

  6. wwe123
    • 3 years ago
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    one min

  7. wwe123
    • 3 years ago
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    AD is the bisector of angle BAC , so angle BAD = angle CAD angle C >angle B angle ADB =180-(ABD+BAD) angle ADC = 180-(CAD+ACD) since angle BAD = angle CAD angle ADB =180-(BAD+ABD) angle ADC =180-(BAD+ACD) angle C >angle B hence, angle ADB > angle ADC.

  8. pratu043
    • 3 years ago
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    Thanks wwe123.

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