Given that P is a point in parallelogram ABCD, line GF passes through point P and is parallel to AB, line EH passes through point P and is parallel to AD. If the area of BFPE =3，the area of PHDG = 5, find the area of ▵APC.

**Area of a triangle?**

**Formula: Area=1/2 base∙height**

It is quite hard to get the length of the three bases. It is even more difficult to get the height on each base. Therefore, this method doesn’t work.**Cut the triangle into parts**Cut the triangle into 2 or 3 smaller triangles makes the situation even more complicated.**Subract from a bigger shape**The area of ▵APC can be subtracted from quadrilateral APCD and ▵ABC. We need to explore now!

**Method 1: Use Triangle ABC as a whole**

**Method 2: use Quadrilateral APCD** **as a whole**

**Key Points**

- The diagonals of a parallelogram cut the parallelogram into
**two congruent triangles.** - Point
**P is not a special point**, it can be anywhere inside of the parallelogram. As long as there is a triangle APC, the location of P is not important. - The magic part of subtraction is that you can
**subtract those unkown parts**or variables in Algebra. So don’t be afraid of those unkowns.

Geometry is art!

A visual representation will open a new door for you.

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