Question

meri help karo...
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Answer 1

Answer:

answer is that question by a u abhi

Answer 2

$\bigstar\rm\blue{GIVEN}\bigstar$

• $\rm\red{AC=CE,AB=AD}$

• $\rm\blue{\angle{BAD}=\angle{EAC}}$

$\bigstar\rm\blue{TO\:PROVE}\bigstar$

• BC=DE.

Now, Taking $\rm\green{\angle{ADC} as\:common\:and\:adding\: to\:\angle{BAD}=\angle{EAC}}$

$\implies\rm{\angle{BAD}+\angle{ADC}=\angle{EAC}+\angle{ADC}}$

$\implies\rm{\angle{BAC}=\angle{EAD}}$....1

Now, in $\rm\blue{\triangle{ABC}\:and\:\triangle{ADE}}$

$\implies\rm{AC=CE(GIVEN)}$

$\implies\rm{\angle{BAC}=\angle{EAD}}$(From..1)

$\implies\rm{AB=AD}$(Given)

Hence,

$\rm\blue{\triangle{ABC}\:and\:\triangle{ADE}}$ Congurent by SAS criteria

So,. BC= DE proved by CPCT.