Math, asked by anne9, 1 year ago

diagonals ac and bd of a trapezium abcd with ab ll dc intersect each other at the point o. using a similarity criterion for two triangles, show that oa\oc =ob\od

Answers

Answered by Vivi842

here is step by step answer of your question

Attachments:

anne9: can u tell me from which similarity criterion ?

Vivi842: as I wrote let OM be parallel to DC so by iaing correspondence /_AMO=/_ADC and /_ AOM=/_ ACD so by usind As criterion ∆AMO is similar to ∆ADC

Vivi842: AA criterion

anne9: thanks

Answered by musicwhispersof

Answer:

Step-by-step explanation:

In triangles AOB and DOC,

Angle AOB = Angle DOC (VERTICALLY OPPOSITE ANGLE)

Angle OBA =ODC(ALTERNATE ANGLES)

Therefore, triangle AOB is similar to triangle DOC

=) OA/OC=OB/OD (CORRESPONDING SIDES OF SIMILAR TRIANGLES)

Hence proved.

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