Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point ‘O’. Using the criterion of similarity for two triangles, show that OA/OC = OB/OD.
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AB||DC----Given In∆AOD AND COB <AOD=<BOC. vertically opposite angles <OAD=<CBO ALT.INT.ANGLES ∆ADO~~∆CBO-- AA AD/CB=DO/BO=AD/OC. CSST
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Answer:
Step-by-step explanation:
In triangle DOC and triangle BOA
angle CDO = angle ABO ( alternate interior angle as AB II CD)
angle DCO = angle BAO ( alternate interior angle as AB II CD)
angle DOC = angle BOA ( V.O.A)
Therefore, triangle DOC is similar to triangle BOA ( AAA similarity criterion)
Therefore DO/BO = OC/OA ( corresponding sides are propotional)
Therefore OA/OC = OB/OD
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