Chapter 5.4, Problem 40WE

a.

To determine

To find: If the given quadrilateral is a special quadrilateral and if then to write the proof.

Answer to Problem 40WE

Yes, the given quadrilateral is a trapezoid

Explanation of Solution

Given:

A quadrilateral with two sides parallel and one diagonal bisects only one angle.

  

$AB$ is parallel to $DC$

  $BD$ is the diagonal that bisects $\angle ADC$

Proof:

A trapezoid is a quadrilateral with only one pair of parallel sides and the other pair of non-parallel sides.

In a trapezium two pairs of adjacent angles formed between the parallel sides and one of the non-parallel side, add up to ${180}^0$ degrees.

Now in quadrilateral $ABCD$ ,

  $AB$ is parallel to $DC$ (given) ….(i).

So, $\angle BAD+\angle ADC={180}^0$ (Angles on the same side of the transversal are supplementary)….(ii).

Therefore by (i) and (ii),

Quadrilateral $ABCD$ is a Trapezium.

Conclusion:

Therefore, quadrilateral $ABCD$ is a Trapezium.

b.

To determine

To find: if the given quadrilateral is a special quadrilateral and if then to write the proof.

Answer to Problem 40WE

Yes, the given quadrilateral is a regular trapezoid

Explanation of Solution

Given:

Quadrilateral with two sides parallel and one diagonal bisects two angles of the quadrilateral.

  

$AB$ is parallel to $DC$

  $BD$ is the diagonal that bisects $\angle ADC$ and $\angle ABC$

Proof:

Since $BD$ bisects $\angle ADC$ ,

So, $\angle ADC=\angle ADB+\angle BDC$ ….(i).

And $\angle ADB=\angle BDC$ ….(ii).

Since $BD$ bisects $\angle ABC$ ,

So, $\angle ABC=\angle ABD+\angle DBC$ ….(iii).

And $\angle ABD=\angle DBC$ ….(iv).

Now since $AB$ is parallel to $DC$ ,

So, $\angle ABD=\angle CDB$ …..(v). (alternate interior angles)

From (ii), (iv), and (v),

  $\angle ADB=\angle BDC=\angle ABD=\angle DBC$

  $\Rightarrow \angle ADC=\angle ABC$ ….(vi)

Now in $\Delta ABD$ and $\Delta CBD$ ,

  $\angle ABD=\angle DBC$ (given)

  $DB=DB$ (common)

  $\angle ADB=\angle BDC$ (given)

  $\therefore \Delta ABD\cong \Delta CBD$ by Angle-Side-Angle (ASA) congruency criteria.

  $\Rightarrow \angle DAB=\angle DCB$ (Corresponding parts of congruent triangles are congruent (equal))….(vii).

Therefore by (vi) and (vii),

The quadrilateral $ABCD$ has two pairs of equal opposite angles and one side of parallel sides (given).

Conclusion:

Therefore, the quadrilateral $ABCD$ is a regular trapezoid.