ABCD is a trapezium diagonals AC and BD intersect at o find the ratio of aod :boc
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ABCD is a trapezium
so,angle ADC = angle BCD
therefore in ∆ADC & ∆BDC
DC=DC(COMMON)
ANGLE ADC = ANGLE BCD
ANGLE DAC =ANGLE CBD(triangle lies between same parallel and same base)
so, triangle ADC = triangla BCD(by aas rule)
∆AOD + ∆ODC = ∆BOC + ∆ODC
so, ∆AOD = ∆ BOC
therefore ratio of both triangle is 1 : 1
so,angle ADC = angle BCD
therefore in ∆ADC & ∆BDC
DC=DC(COMMON)
ANGLE ADC = ANGLE BCD
ANGLE DAC =ANGLE CBD(triangle lies between same parallel and same base)
so, triangle ADC = triangla BCD(by aas rule)
∆AOD + ∆ODC = ∆BOC + ∆ODC
so, ∆AOD = ∆ BOC
therefore ratio of both triangle is 1 : 1
tq bro
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