Question

B ABCD is a parallelogram and angle A< 90°. The height through the vertex D intersects the diagonal AC at point M. The height through vertex B intersects the diagonal AC at point N. Prove that MBND is a parallelogram.

Answer 1

MBND is a parallelogram as BN=MD and BN║MD.

A parallelogram is a quadrilateral whose both pair of opposite side are equal and parallel. Few examples of parallelograms are square, Rhombus  and rectangle.

Given ABCD is a parallelogram and ∠A=90°

Now we have to prove that MBND is a parallelogram

Given BN⊥AC and DM⊥AC

Since BN and DM are both perpendicular to the same straight line AC then we cam say that BN║DM.

Again we know  from the properties of parallelogram that the diagonal divides a parallelogram into two equal triangles.

Therefore the diagonal AC divides the parallelogram into two equal triangles ΔABC and ΔADC.

Now we know that the area of a triangle is given by $\frac{1}{2}$ ×base ×height .

Area of  ΔABC =Area of ΔADC

Since BN⊥AC and DM⊥AC

∴$\frac{1}{2}$×BN×AC=$\frac{1}{2}$×DM×AC

This above equation can be simplified as

BN=MD

Hence BMND is a parallelogram as we know from the properties of parallelogram that when a pair of opposite sides are equal and parallel in a quadrilateral it is a parallelogram.

To learn more about properties of parallelograms:

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