Chapter 14, Problem 26RP

To determine

To prove that the angle bisectors of a kite are concurrent.

Explanation of Solution

Given:

A kite.

Proof:

  

A kite is a quadrilateral with two pairs of adjacent sides equaltherefore, in triangle BCD and triangle BAD

  $\begin{array}{l} BA=BC\left(\text{adjacent sides of a kite}\right)\hfill \\ AD=CD\left(\text{adjacent sides of a kite }\right)\hfill \\ BD\text{ (common})\hfill \end{array}$

  $\begin{array}{l}\Rightarrow △BCD\cong △BAD\hfill \\ \Rightarrow \angle CDB=\angle ADB\hfill \\ \text{and }\angle ABD=\angle CBD\hfill \end{array}$

So,it can be concluded that BD is the angle bisector of angle B and Dto inscribe a circle within a kite

Now, draw the axis of symmetry which is BD in our case bisect one of the non-vertex angles (C or A ) and extend this line so that it meets line BD at any point say o'.

Now, drop a perpendicular from o' to any of the 4 sides of the kite at b' then taking o' as centre and o'b' as radius inscribe a circle within the kiteall kites are tangential quadrilaterals that is they are $4$ sided figures into which a circle can be inscribed such that each side will touch the circle at any one point.