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Dividing Evenly a Quadrilateral: What is this about?
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This is problem 521 (p. 216) from a recently translated to English a famous Russian geometry textbook by A. P. Kiselev (check a review).
| In a quadrilateral ABCD, through the midpoint K of the diagonal BD, the line parallel to the diagonal AC is drawn. Suppose that this line intersects the side AB at a point E. Prove that the line CE bisects the area of the quadrilateral. |
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Triangles ABK and ADK that have the same altitude (from A) and equal bases BK and DK have equal areas. The same holds for triangles CBK and CDK. This implies that quadrilaterals ABCK and ADCK have equal areas.
Triangles AEK and CEK have equal altitudes and the same base, so they also have equal areas. Finally, if L is the intersection of AK and CE, triangles AEL and CKL which are obtained from triangles AEK and CEK by removing triangle LEK, have equal areas.
ΔBCE is obtained from the quadrilateral ABCK by adding ΔCKL and removing ΔAEL. Quadrilateral AECD is obtained from quadrilateral ADCK by removing ΔCKL and adding ΔAEL. The two then have equal areas.
(The proof refers to the configuration described in the problem, i.e. where E lies on AB. The applet however allows for more general configurations. The problem remains valid even for concave quadrilaterals as long as the diagonal BD lies in their interior.)
Remark
Prof. William McWorter informed me of another way of finding point E.
References
- Kiselev's Geometry. Book I. PLANIMETRY, adapted from Russian by Alexander Givental, Sumizdat, 2006.
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