Question 1

We will use triangle congruence to prove that the perpendicular from the center bisects the chord.

Step 1 — Identify given information

Let's consider a circle. It has a center, C. Let AB be a chord of this circle. Let CM be a line segment. CM is drawn from the center C. CM is perpendicular to the chord AB. This means $\angle CMA = 90^\circ$. It also means $\angle CMB = 90^\circ$. We need to show that AM equals BM.

Step 2 — Prove triangle congruence

Let's look at two triangles. These are $\triangle CMA$ and $\triangle CMB$. Side CA is a radius of the circle. Side CB is also a radius of the circle. So, $CA = CB$. Side CM is common to both triangles. So, $CM = CM$. We know $\angle CMA = 90^\circ$. We also know $\angle CMB = 90^\circ$. So, $\angle CMA = \angle CMB$. We can use the RHS congruence rule. RHS stands for Right angle, Hypotenuse, Side. The triangles $\triangle CMA$ and $\triangle CMB$ are congruent. $\triangle CMA \cong \triangle CMB$ By CPCT, corresponding parts of congruent triangles. The side AM corresponds to side BM. So, $AM = BM$. This means point M is the midpoint of chord AB. Therefore, the perpendicular from the center bisects the chord.

Answer

The perpendicular from the center of a circle to a chord bisects the chord because the two triangles formed by the radii, the perpendicular, and the chord segments are congruent by the RHS rule, making the chord segments equal.