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03/06/2021 12:21 pm
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Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O. Using a similarity criterion for two triangles, show that AO/OC = OB/OD.
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03/06/2021 12:23 pm
In ΔDOC and ΔBOA,
AB || CD, thus alternate interior angles will be equal,
∴ ∠CDO = ∠ABO
Similarly,
∠DCO = ∠BAO
Also, for the two triangles ΔDOC and ΔBOA, vertically opposite angles will be equal;
∴∠DOC = ∠BOA
Hence, by AAA similarity criterion,
ΔDOC ~ ΔBOA
Thus, the corresponding sides are proportional.
DO/BO = OC/OA
⇒OA/OC = OB/OD
Hence, proved.
This post was modified 4 years ago by Raavi Tiwari