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त्रिभुज

Question
CBSEHHIMAH10010106

आकृति में एक ही आधार BC पर दो त्रिभुज ABC और DBC बने हुए हैं। यदि AD, BC को O पर प्रतिच्छेद करे, तो दर्शाइए कि fraction numerator ar left parenthesis increment ABC right parenthesis over denominator ar left parenthesis increment DBC right parenthesis end fraction equals AO over DO है।

Solution

A तथा D से BC पर लंब AM और DN खींचो।
∆AMO और ∆DNO में,
space space space space space space space space angle AMO space equals space angle DNO space left parenthesis 90 degree right parenthesis
और space space space space angle AOM space equals space angle DON
        [शीर्षाभिमुख कोण]
      ∆AMO ~ ∆DNO
rightwards double arrow space space space space space space space space space AO over DO equals AM over DN space space space space space space space space space... left parenthesis straight i right parenthesis
अब,  ar left parenthesis increment ABC right parenthesis space equals space 1 half cross times BC cross times AM
और   ar left parenthesis increment BDC right parenthesis space equals space 1 half cross times BC cross times DN
rightwards double arrow space space space fraction numerator ar left parenthesis increment ABC right parenthesis over denominator ar left parenthesis increment BDC right parenthesis end fraction space equals space fraction numerator begin display style 1 half cross times BC cross times AM end style over denominator begin display style 1 half cross times BC cross times DN end style end fraction equals AM over DN
परन्तु          AM over DN equals AO over DO
अत: यह सिद्ध होता है कि
rightwards double arrow   fraction numerator ar left parenthesis increment ABC right parenthesis over denominator ar left parenthesis increment BDC right parenthesis end fraction equals AO over DO 

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