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Question
D, E and F are respectively the mid-points of the sides BC, CA and AB of a ΔABC. Show that
(i) BDEF is a parallelogram.
(ii) ar (DEF) = 1/4ar (ABC)
(iii) ar (BDEF) = 1/2ar (ABC)
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Solution
(i) In ΔABC,
E and F are the mid-points of side AC and AB respectively.
Therefore, EF || BC and EF = 1/2BC (Mid-point theorem)
However, BD = 1/2BC (D is the mid-point of BC)
Therefore, BD = EF and BD || EF
Therefore, BDEF is a parallelogram.
(ii) Using the result obtained above, it can be said that quadrilaterals BDEF, DCEF, AFDE are parallelograms.
We know that diagonal of a parallelogram divides it into two triangles of equal area.
∴Area (ΔBFD) = Area (ΔDEF) (For parallelogram BD)
Area (ΔCDE) = Area (ΔDEF) (For parallelogram DCEF)
Area (ΔAFE) = Area (ΔDEF) (For parallelogram AFDE)
∴Area (ΔAFE) = Area (ΔBFD) = Area (ΔCDE) = Area (ΔDEF)
Also,
Area (ΔAFE) + Area (ΔBDF) + Area (ΔCDE) + Area (ΔDEF) = Area (ΔABC)
⇒ Area (ΔDEF) + Area (ΔDEF) + Area (ΔDEF) + Area (ΔDEF) = Area (ΔABC)
⇒ 4 Area (ΔDEF) = Area (ΔABC)
⇒ Area (ΔDEF) = 1/4Area (ΔABC)
(iii) Area (parallelogram BDEF) = Area (ΔDEF) + Area (ΔBDF)
⇒ Area (parallelogram BDEF) = Area (ΔDEF) + Area (ΔDEF)
⇒ Area (parallelogram BDEF) = 2 Area (ΔDEF)
⇒ Area (parallelogram BDEF)`= 2xx1/4"Area "(ΔABC)`
⇒ Area (parallelogram BDEF) = 1/2Area (ΔABC)
RELATED QUESTIONS
In a triangle ABC, E is the mid-point of median AD. Show that ar (BED) = 1/4ar (ABC).
Show that the diagonals of a parallelogram divide it into four triangles of equal area.
In the given figure, ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F. Show that
(i) ar (ACB) = ar (ACF)
(ii) ar (AEDF) = ar (ABCDE)
ABCD is a trapezium with AB || DC. A line parallel to AC intersects AB at X and BC at Y. Prove that ar (ADX) = ar (ACY).
[Hint: Join CX.]
In the given figure, AP || BQ || CR. Prove that ar (AQC) = ar (PBR).
In the following figure, D and E are two points on BC such that BD = DE = EC. Show that ar (ABD) = ar (ADE) = ar (AEC).
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In the below fig. D and E are two points on BC such that BD = DE = EC. Show that ar
(ΔABD) = ar (ΔADE) = ar (ΔAEC).
ABCD is a parallelogram and X is the mid-point of AB. If ar (AXCD) = 24 cm2, then ar (ABC) = 24 cm2.
The area of the parallelogram ABCD is 90 cm2 (see figure). Find ar (ΔBEF)
In the following figure, ABCDE is any pentagon. BP drawn parallel to AC meets DC produced at P and EQ drawn parallel to AD meets CD produced at Q. Prove that ar (ABCDE) = ar (APQ)
