Diagonals AC and BD of a trapezium ABCD with AB\\DC intersect each other at the point O.Using a simiarity criterion for two triangles Show that OA/OC=OB/OD
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Hey there,
In ΔDOC and ΔBOA,
∠CDO = ∠ABO [Alternate interior angles as AB || CD]
∠DCO = ∠BAO [Alternate interior angles as AB || CD]
∠DOC = ∠BOA [Vertically opposite angles]
∴ ΔDOC ~ ΔBOA [AAA similarity criterion]
∴ DO/BO = OC/OA [ Corresponding sides are proportional]
⇒ OA/OC = OB/OD
Hope this helps!
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After the proves of AAA theorem any other steps to follow
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