Chapter 5.5, Problem 28WE

To determine

To discover, state and prove all about the diagonals and angles of a kite.

Explanation of Solution

Given:

A convex kite

  

Concept used:

A kite is a quadrilateral that has two pairs of congruent sides, but opposite sides are not equal.

A kite $ABCD$ is given.

1. The diagonals of a kite are perpendicular.

Given:

Kite $ABCD$

  

To Prove: $\overrightarrow{BD}\perp \overrightarrow{AC}$

Proof:

A kite has two distinct sets of congruent adjacent sides.

So,

  $\overrightarrow{AB}\cong \overrightarrow{BC}$ and $\overrightarrow{AD}\cong \overrightarrow{CD}$ ….(i).

Now, in $\Delta BAD$ and $\Delta BCD$

  $\overrightarrow{AB}=\overrightarrow{BC}$ [from (i)]

  $\overrightarrow{AD}=\overrightarrow{CD}$ [from (i)]

  $\overrightarrow{BD}=\overrightarrow{BD}$ [common side]

Hence, by Sid-Side-Side (SSS) congruency criteria $\Delta BAD\cong \Delta BCD$ .

  $\Rightarrow \angle ABD=\angle CBD$ [ $\because$ Corresponding parts of congruent triangles are congruent (equal)]

Now, in $\Delta BEA$ and $\Delta BCE$

  $\overrightarrow{AB}=\overrightarrow{BC}$ [from (i)]

  $\angle ABE=\angle CBE$ (Corresponding parts of congruent triangles)

  $\overrightarrow{BE}=\overrightarrow{BE}$ [common side]

Hence, by Sid-Angle-Side (SAS) congruency criteria $\Delta BEA\cong \Delta BCE$ .

  $\Rightarrow \angle BEA=\angle BEC$ [ $\because$ Corresponding parts of congruent triangles are congruent (equal)]

Now, $\angle BEA+\angle BEC={180}^0$ [ Linear Pair]

  $\Rightarrow \angle BEA+\angle BEA={180}^0$ [ $\because \angle BEA=\angle BEC$ ]

  $\Rightarrow 2\angle BEA={180}^0$

  $\Rightarrow \angle BEA={90}^0$

So, $\angle BEA=\angle BEC={90}^0$

Hence, diagonals of a kite are perpendicular to each other.

Hence proved.

2. The diagonal of a kite bisects the pair of opposite angles.

Given:

A kite $ABCD$

  

To prove:

  $\overrightarrow{BD}$ bisects $\angle ABC$

  $\overrightarrow{BD}$ bisects $\angle ADC$

Proof:

A kite has two distinct sets of congruent adjacent sides.

So,

  $\overrightarrow{AB}\cong \overrightarrow{BC}$ and $\overrightarrow{AD}\cong \overrightarrow{CD}$ ...(i).

Now, in $\Delta BAD$ and $\Delta BCD$

  $\overrightarrow{AB}=\overrightarrow{BC}$ [from (i)]

  $\overrightarrow{AD}=\overrightarrow{CD}$ [from (i)]

  $\overrightarrow{BD}=\overrightarrow{BD}$ [common side]

Hence, by Sid-Side-Side (SSS) congruency criteria $\Delta BAD\cong \Delta BCD$ .

  $\Rightarrow \angle ABD=\angle CBD$ [ $\because$ Corresponding parts of congruent triangles are congruent (equal)]

  $\Rightarrow \angle ADB=\angle CDB$

Therefore, $\overrightarrow{BD}$ bisects $\angle ABC$ and $\overrightarrow{BD}$ bisects $\angle ADC$

Hence proved.

Conclusion:

Diagonals are perpendicular to each other and bisects opposite angles.