Question

In a quadrilateral ABCD, If bisectors of the $\angle ABC$ and $\angle ADC$ meet on the diagonal AC, prove that the bisectors of $\angle BAD$ and $\angle BCD$ will meet on the diagonal BD.

Answer 1

Given ABCD is a quadrilateral in which the bisectors of $\angle ABC$ and $\angle ADC$ meet on the diagonal AC at P.

To proove Bisectors of $\angle BAD$ and $\angle BCD$ meet on the diagonal BD.

Construction Join BP and DP. LEt the bisector of $\angle BAD$ meet BD at Q. Join AQ and CQ.

Proof In order to prove that the bisectors of $\angle BAD$ and $\angle BCD$ meet on the diagonal BD. It is sufficient to prove that CQ is the bisector of $\angle BCD$. For which we will prove that Q divides BD in the ratio BC:DC.

In $△ABC$, BP is the bisector of $\angle ABC$.

$\therefore$ $\frac{AB}{BC}=\frac{AP}{PC}$.......(i)

In $△ACD$, DP is the bisector of $\angle ADC$.

$\therefore$ $\frac{AD}{DC}=\frac{AP}{PC}$.......(ii)

From (i) and (ii), we get

$\frac{AB}{BC}=\frac{AD}{DC}$

$\Rightarrow$ $\frac{AB}{AD}=\frac{BC}{DC}$........(iii)

In $△ABD$, AQ is the bisector of $\angle BAD$ [By construction]

$\Rightarrow$ $\frac{AB}{AD}=\frac{BQ}{DQ}$

From (iii) and (iv), we get

$\Rightarrow$ $\frac{BC}{DC}=\frac{BQ}{DQ}$.

Thus, in $△CBD$, Q divides BD in the ratio CB:CD. Therefore, CQ is the bisectors of $\angle BCD$
Hence, bisectors of $\angle BAD$ and $\angle BCD$ meet on the diagonal BD.