P is any point on the diagonal AC of a parallelogram ABCD. Prove that ar(∆ADP) = ar(∆ABP).
Join BD.
Let BD and AC intersect at point O.
O is thus the midpoint of DB and AC.
PO is the median of
So,
$\operatorname{ar}(\triangle \mathrm{DPO})=\operatorname{ar}(\triangle \mathrm{BPO}) \quad \ldots(1)$
$\operatorname{ar}(\triangle \mathrm{ADO})=\operatorname{ar}(\triangle \mathrm{ABO}) \quad \ldots . .(2)$
Case 1 .
$(2)-(1)$
$\Rightarrow \operatorname{ar}(\triangle \mathrm{ADO})-\operatorname{ar}(\triangle \mathrm{DPO})=\operatorname{ar}(\triangle \mathrm{ABO})-\operatorname{ar}(\triangle \mathrm{BPO})$
Thus, ar(∆ADP) = ar(∆ABP)
Case II:
$\operatorname{ar}(\triangle \mathrm{ADO})+\operatorname{ar}(\triangle \mathrm{DPO})=\operatorname{ar}(\triangle \mathrm{ABO})+\operatorname{ar}(\triangle \mathrm{BPO})$
Thus, $\operatorname{ar}(\Delta \mathrm{ADP})=\operatorname{ar}(\Delta \mathrm{ABP})$
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