Question

In the given figure ABCD is a parallelogram. If P is a point on the diagonal, which of the following is true?

• A. $ar\left(△APD\right)=ar\left(△BPC\right)$
• B. $ar\left(△CPD\right)=ar\left(△APB\right)$
• C. $ar\left(△APD\right)=ar\left(△APB\right)$
• D. $ar\left(△APD\right)=ar\left(△PDC\right)$

Answer 1

The correct option is C $ar\left(△APD\right)=ar\left(△APB\right)$

Let’s draw perpendicular from B and D to the diagonal.
Take $△ADQ$ and $△CBR$
$\angle Q=\angle R\left(each{90}^{\circ }\right)$
$\angle A=\angle C$ ( alternate interior angles)
So, $\angle D=\angle B$ (third angle of triangle)
AD = BC ( opposite sides of parallelogram)
Using ASA criteria
$△ADQ\cong △CBR$
So, $DQ=BR$
$ar\left(△APD\right)=\frac{1}{2}\times AP\times DQ$
$ar\left(△APB\right)=\frac{1}{2}\times AP\times BR$
We know that DQ = BR
Hence $ar\left(△APD\right)=ar\left(△APB\right)$