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).

• A. True
• B. False

Answer 1

The correct option is A True

Between $\Delta$ AOD & $\Delta$ BOC we have AO=CO (given),BO=OD (by Construction).

$\angle$ AOD $=\angle$ BOC....Vertically opposite angle

$\therefore$ By SAS test $\Delta$ AOD & $\Delta$ BOC are congruent.

So AD=BC....(i)

similarly, between $\Delta$ AOB & $\Delta$ DOC we have AO=CO (given), BO=OD (by Construction)

$\angle$ AOB $=\angle$ DOC

$\therefore$ By SAS test $\Delta$ AOB & $\Delta$ DOC are congruent.

So AB=DC.....(ii)

Also $\angle$ ABC$={90}^o$....(iii)

$\therefore$ From (i) & (ii) & (iii) we conclude that ABCD is a rectangle.

So the diagonals AC & BD are equal and bisect each other at O.

$\therefore$ OA=OB=OC=OD.

i.e O is equidistant from A, B & C.