Question

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Answer 1

Answer:

In Δs AED,BED ,

AE=BE

DE is common

included angles DEA = DEB=90∘ (Construction)

So, ΔAED=ΔBED

Therefore, BD=AD=AC2

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

=> AC = 2BD

=> BD = AC/2

[Hence Proved]

Answer 2

Step-by-step explanation:

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