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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  $$ar( A B C ) = ar ( A B D ).$$
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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 areas

Similarly, 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 areas

Adding eqns$$(2)$$ and $$(3)$$ we get

ar$$\left(\triangle{AOC}\right)+$$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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