Question

can someone help me with the 19th ques 2choice

Answer 1

here's your answer,

go for it..

Answer 2

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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