Question

in a triangle ABC
PQ II BC
P be the mid point of side AB and Q be the mid point of AC
then prove that area of triangle OBC=area of rectangle OPAQ

[note-- QB and PC are crossing lines inside the triangle and they meet at a point O which lies inside the
triangle

Answer 1

Answer on Question #45243 – Math - Geometry

in a triangle ABC PQ II BC

P be the mid point of side AB and Q be the mid point of AC

then prove that area of triangle OBC=area of rectangle OPAQ

[note-- QB and PC are crossing lines inside the triangle and they meet at a point O which lies inside the triangle]

If P is midpoint of AB than CP is median,

if Q is midpoint of AC than BQ is median.

Let AR be the third median.

O is the centroid (point where medians meet).

One of the properties of medians is that the three medians divide the triangle into 6 smaller triangles that all have the same area, even though they may have different shapes.

I.e. area of $OPA =$ area of $OQA =$ area of $OPB =$ area of $OQC =$

$$= \text{area of } OBR = \text{area of } ORC = \frac{1}{6} \text{ (area of } ABC\text{)}$$

Area of $OPQA =$ area of $OPA +$ area of $OQA$

Area of $OBC =$ area of $OBR +$ area of $ORC$

Therefore:

Area of $OPQA =$ area of $OBC = \frac{1}{3}$ (area of $ABC$).

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