  Question

$$ABC$$  and  $$A B D$$  are two triangles on the same base  $$A B.$$  If line-segment  $$CD$$  is bisected by  $$AB$$  at  $$O,$$  show that  $$ar( A B C ) = ar ( A B D ).$$ Solution

Given:$$\triangle{ABC}$$ and $$\triangle{ABD}$$ on the same base $$AB$$ and $$AB$$ bisects $$CD$$$$OC=OD$$    .....$$(1)$$To prove:ar$$\left(\triangle{ABC}\right)=$$ar$$\left(\triangle{ABD}\right)$$Proof:In $$\triangle{ACD},$$Since $$OC=OD$$    from $$(1)$$$$\therefore,OA$$ is the median$$\Rightarrow$$ ar$$\left(\triangle{AOC}\right)=$$ar$$\left(\triangle{AOD}\right)$$   ....$$(2)$$Since median divides the triangle into equal areasSimilarly, in $$\triangle{BCD}$$since $$OC=OD$$   from $$(1)$$$$\therefore OB$$ is the median$$\Rightarrow$$ ar$$\left(\triangle{BOC}\right)=$$ar$$\left(\triangle{BOD}\right)$$   ....$$(3)$$Since median divides the triangle into equal areasAdding eqns$$(2)$$ and $$(3)$$ we getar$$\left(\triangle{AOC}\right)+$$ar$$\left(\triangle{BOC}\right)=$$ar$$\left(\triangle{AOD}\right)+$$ar$$\left(\triangle{BOD}\right)$$$$\Rightarrow$$ar$$\left(\triangle{ABC}\right)=$$ar$$\left(\triangle{ADB}\right)$$Hence proved.Mathematics

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