Chapter 5.7, Problem 10PSA

To determine

To prove: That the quadrilateral $BIRD$ is a kite.

Explanation of Solution

Given Information:

  $ID$ bisects $RB$

  $BI=IR$

Formula used:

Corresponding parts of congruent triangles are congruent

Proof:

In the quadrilateral $BIRD,ID$ bisects $BR$

  $\Rightarrow BK=KR\mathrm{...}\left(\text{i}\right)$

  $BI=IR$ … (Given)

  $IK=IK$ … (Common side)

  $\Rightarrow \Delta BKI\cong \Delta RKI$ … (By SSS congruency test)

As we know that the corresponding parts of congruent triangles are congruent

  $\Rightarrow \angle BKI=\angle RKI$ … (ii)

As $BI$ is a straight line

  $\begin{array}{l}\Rightarrow \angle BKR={180}^{\text{o}}\\ \Rightarrow \angle BKI+\angle IKR={180}^{\text{o}}\end{array}$

  $\Rightarrow \angle BKI=RKI={90}^{\text{o}}$ … From (ii)

As $\angle BKI+\angle BKD={180}^{\text{o}}$

  $\begin{array}{l}\Rightarrow {90}^{\text{o}}+\angle BKD={180}^{\text{o}}\\ \Rightarrow \angle BKD={90}^{\text{o}}\end{array}$

Similarly, $\angle RKD={90}^{\text{o}}$

In $\Delta BKD\text{and}\Delta RKD,$

  $\angle BKD=\angle RKD$ … (Both right angles)

  $\begin{array}{l}BK=KR\mathrm{...}\left(\text{Given}\right)\\ KD=KD\mathrm{...}\left(\text{Common}\text{side}\right)\end{array}$

  $\therefore \Delta BKD\cong RKD$ … (By SAS congruency rule)

As the corresponding parts of congruent triangles are congruent

  $\Rightarrow BD=RD$

Hence, $BIRD$ is a kite.