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In figure (1) given below, point D divides the side BC of ∆ABC in the ratio m : n. Prove that area of ∆ABD : area of ∆ADC = m : n.

In figure (1) given below, point D divides the side BC of ∆ABC in the ratio m : n. Prove that area of ∆ABD : area of ∆ADC = m : n. Theorems on Area, ML Aggarwal Understanding Mathematics Solutions ICSE Class 9.

Theorems on Area

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Answer

From fig (1)

In figure (1) given below, point D divides the side BC of ∆ABC in the ratio m : n. Prove that area of ∆ABD : area of ∆ADC = m : n. Theorems on Area, ML Aggarwal Understanding Mathematics Solutions ICSE Class 9.

In ∆ABC, point D divides the side BC in the ratio m : n.

BD : DC = m : n

area of ∆ABD = 12\dfrac{1}{2} × base × height

= 12\dfrac{1}{2} × BD × AE …….. (i)

area of ∆ADC = 12\dfrac{1}{2} × DC × AE …… (ii)

Dividing (i) by (ii)

⇒area of ∆ABDarea of ∆ADC=12×BD×AE12×DC×AE⇒area of ∆ABDarea of ∆ADC=BDDC⇒area of ∆ABDarea of ∆ADC=mn.\Rightarrow \dfrac{\text{area of ∆ABD}}{\text{area of ∆ADC}} = \dfrac{\dfrac{1}{2} × BD × AE}{\dfrac{1}{2} × DC × AE} \\[1em] \Rightarrow \dfrac{\text{area of ∆ABD}}{\text{area of ∆ADC}} = \dfrac{BD}{DC} \\[1em] \Rightarrow \dfrac{\text{area of ∆ABD}}{\text{area of ∆ADC}} = \dfrac{m}{n}.

Hence, proved that area of ∆ABD : area of ∆ADC = m : n.

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