Question

pls answer it fast........​

Answer 1

Answer:

In △ADC,

$AC^{2} =AD^{2} +DC^{2} --(1_$)

in triangle ABD$AB^{2} =AD^{2} +DB^{2}$

Therefore $AD^{2} =AB^{2} -DB^{2}$-----(1)

Substituting values in eq(1)

$AC^{2}$ = $(AB^{2}-DB ^{2} ) +(DB+BC)^{2}$

       =$BC^{2}-DB^{2} +DB^{2} +BC^{2}+2DB.BC$     (As, AB=BC)

        =$2BC^{2} +2DB.BC$

       =2BC(BC+DB)

$AC^{2}$  =2BC×DC

hence proved...

Answer 2

Step-by-step explanation:

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