Question

in the given figure ABC is a right triangle in which angle B = 90° . if O is the mid - point of hypotenuse AC, prove that BO = 1/2 AC

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Answer 1

GIVEN: A right triangle ABC, right angled at B. D is the mid point of AC, ie, AD = CD

TO PROVE : BD = AC/2

Since, circumcentre of any right triangle is the mid point of its hypotenuse.

=> D is the centre of the circle passing through A, B & C

=> AD = BD = CD ( being radii of the same circle)

Or

2AD = 2BD

2AD = 2BD=> AC = 2BD

2AD = 2BD=> AC = 2BD=> BD = AC/2

[Hence Proved]

Answer 2

Answer:

Given

∆ABC is a right triangle

angle =90°

O is the mid point of AC

AO = CO

construct a imaginary line DC parallel to AB

O as mid point of BD

AO=CO

BO=DO

In ∆ABO , ∆DOC

AO=CO

angle AOB = angle DOC ( vertically opposite angle )

BO=DO

∆ABO,congrence ∆DCO

AB=DC, angle BAO= angle CDO, angle ABO =angle DCO[by C.P.C.T]

to prove BO=1/2AC

In∆ABC,∆DCB

AB=DC

Angle ABC=90°

angle ABC + angle DCO = 180°

90°+angle DCO =180°

angle DCO =180°-90°

angle DCO=90°

BC=BC

∆∆ABC congrence ∆DCB

BD=AC

1/2BD =1/2AC

BO=1/2AC