Question

ABC is a right-angled triangle and O is the midpoint of the side opposite to the right angle. Explain why O is equidistant from A, B and C.
(The dotted lines are drawn additionally to help you)

Answer 1

Hint: First we will draw a line from A parallel to BC and a line from C parallel to BA. Then we will let them meet at point D and join OD. Then we will prove ABCD is parallelogram as opposite sides are parallel and then prove this parallelogram to be a rectangle by using the basic properties of a rectangle. We will use the property of a rectangle that the diagonal of a rectangle bisect other and are of equal length to prove.

Complete step by step solution: We are given that ABC is a right-angled triangle and O is the midpoint of the side opposite to the right angle.

First, we will draw a line from A parallel to BC and a line from C parallel to BA.
Then we will let them meet at point D and join OD.

First, we will take ABCD, where we know that $$\text{AB||DC}$$ and $$\text{BC||AD}$$ by our above construction.
Thus, the opposite sides are parallel.
Hence, ABCD is a parallelogram.

We also know that adjacent angles of a parallelogram are supplementary, from the above diagram, the sum of angle $$\mathrm{\angle }ABC$$ and $$\mathrm{\angle }BCD$$ is $${180}^{\circ }$$, we have
$$\Rightarrow \mathrm{\angle }ABC+\mathrm{\angle }BCD={180}^{\circ }$$

Substituting the value of $$\mathrm{\angle }ABC$$ in the above equation, we get
$$\Rightarrow {90}^{\circ }+\mathrm{\angle }BCD={180}^{\circ }$$

Subtracting both sides by $${90}^{\circ }$$ in the above equation, we get
$$\begin{gathered} \Rightarrow {90}^{\circ }+\mathrm{\angle }BCD-{90}^{\circ }={180}^{\circ }-{90}^{\circ } \\ \Rightarrow \mathrm{\angle }BCD={90}^{\circ } \end{gathered}$$

Also, we know that the opposite angles of a parallelogram are equal, so we have
$$\begin{gathered} \Rightarrow \mathrm{\angle }DAB=\mathrm{\angle }BCD \\ \Rightarrow \mathrm{\angle }DAB={90}^{\circ } \end{gathered}$$

$$\begin{gathered} \Rightarrow \mathrm{\angle }ADC=\mathrm{\angle }ABC \\ \Rightarrow \mathrm{\angle }ADC={90}^{\circ } \end{gathered}$$

Therefore, each angle of the parallelogram ABCD is a right angle.

So we know that when a parallelogram with all right angles is a rectangle.

Thus, ABCD is a rectangle.

We also know that the diagonals of a rectangle bisect each other, then we have from the above diagram that
$$\Rightarrow OA=OC={\displaystyle \frac{1}{2}}AC\text{ ......eq.(1)}$$
$$\Rightarrow OB=OD={\displaystyle \frac{1}{2}}BD\text{ ......eq.(2)}$$

We also know that the diagonal of a rectangle are equal length, then we have
$$\Rightarrow BD=AC$$

Dividing both sides by 2 in the above equation, we get
$$\Rightarrow {\displaystyle \frac{1}{2}}BD={\displaystyle \frac{1}{2}}AC$$

Using equation (1) and equation (2) in the above equation, we get
$$\Rightarrow OB=OA$$

Therefore, we have found out that $$OB=OA=OC$$.
Hence, O is equidistant from A, B and C.
Hence, proved.

Note: In solving these types of questions, you need to know that the properties of rectangles and their diagonals. Then we will use the properties accordingly. This is a simple problem, one should only need to know the definitions. It is clear from the diagram that it is a rectangle as nowhere it is given it to be a square, so remember that as well.