Question

100 points with explanation pls answer

Answer 1

I am not using the hint to solve it.

We know that any triangle inscribed in a semicircle is a right angled triangle.
Here, the diameter of the circle is the hypotenuse of the triangle.

Coming to the question, Angle B is 90°. So AC is the hypotenuse of the triangle.

If you construct a semicircle with AC as diameter, you will see that vertex B will lie on the semicircle.

Given that point O is midpoint of AC.

So OA = OC = AC/2 = radius of semicircle

Join O and B.
Since B lies in semicircle, OB is also a radius

Thus, OA = OB = OC = AC/2

(Proved)

Answer 2

$\underline{\bold{Given:-}}$

∆ABC is a triangle right angled at B. O is the midpoint of AC.

$\underline{\bold{To\:prove:-}}$

$OA \: = OB \: = OC\: = \frac{AC}{2}$
$\underline{\bold{Construction:-}}$

Draw l parallel to BC intersecting AB at D.

$\underline{\bold{Proof:-}}$

Since, O is the midpoint if AC.

So, OA = OC = AC/2 ...........(1)

Using converse of midpoint theorem ,

The line drawn through the midpoint of one side of a triangle parallel to another side bisect the third side.

So, AD = BD

In ∆ ADO and ∆ BDO

OD = OD (common side)

angle ADO = angle BDO (right angles)

AD = BD (proved above)

So by SAS congruency , ∆ ADO is congruent to ∆ BDO.

Therefore,

OA = OB (c.p.c.t.) ......(2)

from eq (1) and (2)

$OA \: = OB \: = OC\: = \frac{AC}{2}$

Hence proved