can someone help me with the 19th ques 2choice
Answers
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Answer:
Step-by-step explanation:
By Pythagoras theorem
In triangle AEC
(AE)^2 + (EC)^2 = (AC)^2
(AE)^2 = (AC)^2 -(EC)^2. -equation(1)
In triangle AED
(AE)^2 + (ED)^2 = (AD)^2
AE)^2 = (AD)^2 - (ED)^2 - equation (2)
Equate equation (1) and (2)
EC = ED + DC
(AC)^2 - (EC)^2 = (AD)^2 - (ED)^2
(AC)^2 - (ed + dc)^2 = ad square - ed square
AC square -( (ed)^2 + (dc)^2 +2 ed.dc)
= Ad square - ed square
AC square - ed square - dc square -2 ed.dc = ad square - ed square
(AC)^2 - (dc)^2 - 2ed.dc = ad square
Ac^2 = AD^2 + DC^2 +2ED.DC
Be the 3rd equation
In triangle AEB
(AB)^2 = (AE)^2 + (BE)^2
Be the 4th equation
Add 3rd and 4th equations
Ac^2 + Ab^2 = AD^2 + dc^2 + 2ed.dc + Ae^2 + Be^2
We have proved LHS
Rhs=
We know that dc= BD ,since d is the midpoint on BC
Rhs = ad^2 + bd^2 +2(ed)(BD) + ae^2 + be square
Be = ed =BD/2
Since 'e' is the mid point
Substitute 'ae^2' as AD^2 - ED^2
=AD^2 + Bd^2 +2(BD/2)(Bd) + AD^2 -ED^2 + BE^2
Ed = be
Hence be square - ed square = 0
BC = 2BD
= 2AD^2 + BD^2 + (BD)(BD)
= 2AD^2 + 2BD^2
= 2(AD^2+ BD^2)
We have proved RHS also
Hope this helps you
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