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Quadrilateral problem in junior two mathematics
(1) Because the parallelogram ABCD

So AB//CD AB=CD

Because the two midpoints

So AE = be =1/2abdf =1/cd-be = df.

So the quadrilateral DEBF is a parallelogram (a set of opposite sides are parallel and equal).

And because AD= 1/2 AB.

So AD=AE

Because the angle DAB = 60

So the triangle ADE is an equilateral triangle.

So DE=AE=BE

So the quadrilateral DEBF is a rhombus (parallelogram with equal adjacent sides).

(2) The quadrilateral AGBD is rectangular because:

Because AD//CG, AG//BD

So the quadrilateral AGBD is a parallelogram (two groups of opposite sides are parallel respectively).

Because DE=BE, AE=DE

So angle ADE= angle DAB, angle BDE= angle ABD.

So angle ADE+ angle BDE= 1/2 (angle ADE+ angle BDE+ angle DAB+ angle ABD)=90.

The quadrilateral AGBD is a rectangle (a parallelogram with right angles).