Question

in the given figure ABCD is a quadrilateral b p is drawn parallel to AC and B p meets DC at P prove that area triangle ADP equal to area of quadrilateral ABCD​

Answer 1

$area (ABCD) = Area (\triangle ABC) + area(\triangle DAC)$

$\triangle DAC \:and \:\triangle PAC \:lie \:on \\the \:same \:base \: \overline {AC} \:and \\ between\:the \: parallels \:DP \parallel AC$

$area(\triangle DAC) = area(\triangle PAC)$

$\pink { ( Triangles \:on \:the \:same \:base}\\\pink{ and \: between \:the \:same \: parallels \:are}\\\pink{ equal \: in \:area ) }$

/* Adding areas of same figures on both sides .*/

$area(\triangle DAC) + area(\triangle ABC)= area(\triangle PAC) + area(\triangle ABC)$

$Hence, area(\triangle ABCD) = area(\triangle ABP)$

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