Consider the quadrilateral ABCD.
draw a diagonal AC.
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Step-by-step explanation:
Given: ABCD is a parallelogram. BK and DL are perpendiculars drawn to diagonals AC.
It is given that in quadrilateral ABCD, DLand BK are perpendicular to AC.
ΔADC is congruent to ΔABC (Diagonal divides a parallelogram into two congruent triangles)
Therefore, Area ΔABC= Area ΔADC
That is 21AC×BK=21AC×DL implies BK=DL.
Hence, BK=DL.
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