Question

$\text{AB}$ and $\text{AC}$ are two equal chords of a circle. Prove that the bisector of the angle $\text{BAC}$ passes through the centre of the circle.

Answer 1

Triangel congruency:

According to the given details

$\text{AB}$ and $\text{AC}$ are two chords that are equal with centre $\text{O}$

$\text{AM}$ is the bisector of $\angle \text{BAC}$

Now join $\text{BC}$ .

In $△\text{BAP}$ and $△\text{CAP}$

$$\begin{gathered} \text{AB=AC(Given)} \\ \angle \text{BAP=∠CAP}\mathbf{\text{(angle bisector)}} \end{gathered}$$

$$\begin{gathered} \text{AP=AP} \\ \text{△BAP≅△CAP(By SAS)} \\ \angle \text{BPA=∠CPA (CPCT)} \end{gathered}$$

We know that,

$\text{CP=PB}$

Since $\angle \text{BPA}$ and $\angle \text{CPA}$ are linear pair angles,

$$\begin{gathered} \angle \text{BPA+∠CPA=180°} \\ \angle \text{BPA=∠CPA=90°} \end{gathered}$$

Then,

AP is the perpendicular bisector of chord $\text{BC}$,which will pass through the centre $\text{O}$ on being produced .

Therefore, it is proved that $\text{AM}$ passes through the centre $\text{O}$ .