Question

if ED||AB than prove that 1+AD/CD=1+EB/CE​

Answer 1

Answer:

AEX EB = CE - ED ( Proved )

= Step - by - step explanation :

Now , in A CEB and A AED ,

We have to prove that ,

AE × EB = CE × ED Z CEB = < AED ( They are vertically opposite angles } < BCE = < EAD ( Those are the angles inscribed in the same arc } So ,

\ frac { CE } { AE } = \ frac { BE } { ED } AE CE ~

So , A CEB A AED { A - A test of similarity}

|| So , \ frac { CE } { AE } = \ frac { BE } { ED } AE CE 82 ED BE

{ The corresponding sides of two similar triangles will be in the same ratio }

Step-by-step explanation:

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Answer 2

Step-by-step explanation:

ED||AB (Given)

$\frac{CE}{CB} =\frac{CD}{CA}$ (Line parallel to one side of a triangle divides its other two

                sides in the same ratio)

=>  $\frac{CB}{CE}=\frac{CA}{CD}$

=> $\frac{CE+EB}{CE} =\frac{CD+AD}{CD}$

=> $\frac{CE}{CE}+\frac{EB}{CE} =\frac{CD}{CD} +\frac{AD}{CD}$

=> $1+\frac{EB}{CE} =1+\frac{AD}{CD}$

=> $1+\frac{AD}{CD}=1+\frac{EB}{CE}$

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