Question

ABC is a right-angled triangle and O is the mid point of the side opposite to the right angle. Explain why O is equidistant from A, B and C.


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Answer 1

Answer:

yes o is equidistant from A , B and C

Explanation:

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

∠ AOD =∠ BOC....Vertically opposite angle

∴ By SAS test Δ AOD & Δ BOC are congruent.

So AD=BC....(i)

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

∠ AOB =∠ DOC

∴ By SAS test Δ AOB & Δ DOC are congruent.

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

Also ∠ ABC=90

o

....(iii)

∴ From (i) & (ii) & (iii) we conclude that ABCD is a rectangle.

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

∴ OA=OB=OC=OD.

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

Answer 2

Answer:

Opposite sides are parallel and equal to each other and all the interior angles are right angles. The property of rectangle states that the diagonals are of equal length and bisect each other. Hence, AO = OC = BO = OD. Thus, O is equidistant from A, B and C.

Explanation:

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