If P is any point on the median AD of a trian | vedclass

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(A) The statement is True.$1$. In $	riangle ABC$,$AD$ is the median,so it divides the triangle into two triangles of equal area. Thus,$\operatorname{

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(A) The statement is True.$1$. In $	riangle ABC$,$AD$ is the median,so it divides the triangle into two triangles of equal area. Thus,$\operatorname{ar}(ABD) = \operatorname{ar}(ACD)$.$2$. Similarly,in $	riangle PBC$,$PD$ is the median,so $\operatorname{ar}(PBD) = \operatorname{ar}(PCD)$.$3$. Subtracting the area of $	riangle PBD$ from $	riangle ABD$ and $	riangle PCD$ from $	riangle ACD$,we get $\operatorname{ar}(ABD) - \operatorname{ar}(PBD) = \operatorname{ar}(ACD) - \operatorname{ar}(PCD)$.$4$. This simplifies to $\operatorname{ar}(ABP) = \operatorname{ar}(ACP)$.

If $P$ is any point on the median $AD$ of a $\triangle ABC$,then $\operatorname{ar}(ABP) = \operatorname{ar}(ACP)$. State whether this statement is True or False.

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