someone please answer this question
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hello friend , here is your answer ,
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In ∆ ACE and ∆ BED
angle A = angle D ( corresponding
angle C = angle B. angles. )
therefore , by AA criterion ,
∆ ACE ~ ∆ BED
thus , by Thales theorem ,
![\frac{ae}{ce} = \frac{de}{be} \frac{ae}{ce} = \frac{de}{be}](https://tex.z-dn.net/?f=+%5Cfrac%7Bae%7D%7Bce%7D++%3D++%5Cfrac%7Bde%7D%7Bbe%7D+)
____________________
hope my answer helped you
_______________________
In ∆ ACE and ∆ BED
angle A = angle D ( corresponding
angle C = angle B. angles. )
therefore , by AA criterion ,
∆ ACE ~ ∆ BED
thus , by Thales theorem ,
____________________
hope my answer helped you
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