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Byju's Answer

Standard IX

Mathematics

The Mid-Point Theorem

Figure shows ...

Question

Figure shows that AD and BF are medians of ∆ABC and BF || DE then CE is equal to

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Solution

The correct option is B

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Similar questions

Q. In the given figure,

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Q. In the given figure, DE and BF are the bisectors of ∠ADC and ∠CBA respectively. If DE = BF, AD || BC, AD = BC, AE = 4 cm and BE = 6 cm, then FC + DF equals

दिए गए चित्र में, DE तथा BF क्रमशः ∠ADC तथा ∠CBA के समद्विभाजक हैं। यदि DE = BF, AD || BC, AD = BC, AE = 4 cm तथा BE = 6 cm है, तब FC + DF बराबर है

In the given figure, AD, CE and BF are the medians. O is the point at which the medians of the $Δ A B C$ intersect. Area of the $Δ B O C$ is $24 c m^{2}$ . Then the area of the $Δ A B C$ is:

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