In an isosceles triangle ABC with …
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Sia ? 5 years, 5 months ago
In {tex}\triangle {/tex}ABC, BD {tex} \bot {/tex} AC
{tex}\therefore {/tex} In right {tex}\triangle {/tex} ADB, BD2 = AB2 - AD2 ...........(i)
In right {tex}\triangle {/tex} BDC, CD2=BC2-BD2 .............(ii)
Subtracting (ii) from (i) we get
BD2 - CD2 = AB2 - AD2 - BC2 + BD2
= AC2 - AD2 - BC2 + BD2
= (AD + CD)2 - AD2 - BC2 + BD2 [AB = AC, Given]
=(CD2 + BD2) - BC2 + 2AD.CD [AC = AD + CD]
= BC2 - BC2 + AD.CD [CD2+BD2 = BC2]
=2AD.CD Proved
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