Question

question is in attachment​

Answer 1

In the given figure, length of AB is equal to the length of AC .

Due to the property of isosceles triangle, $\angle$ABC = $\angle$ ACB,

For convenience, let $\angle$ ABC =$\angle$ = 2y

In ∆ABD and ∆ADC,

= > $\angle$ ADB =$\angle$ADC ,$\angle$ABC= $\angle$ACB, AD = AD

So, ∆ABD $\cong$∆ADC

Thus,

$\angle$BAD =$\angle$CAD

Let $\angle$BAD =$\angle$CAD = x

Now,

Let the intersection of AD and EC be O,

Then,

$\angle$ AOE = $\angle$ COD [vertically opposite angles ]

= > 180 - ( 90 - x ) = 180 - ( 90 - y ) [ $\angle$ COD = y, as ∆AEC $\cong$ BEC, which can be proved ]

= > 90 - x = 90 - y

= > x = y

= > 2x = 2y

= > $\angle$ BAC = $\angle$ ABC

So,

$\angle$ BAC = $\angle$ ABC

$\angle$ AEC = $\angle$ ADB = 90°

$\angle$ ACE or y = $\angle$ BAD or x or y

Therefore, ∆ABD $\cong$∆CAE

So, AD = EC by cpct .

$\:$

Answer 2

Answer:

Hey mate, the answer is in attachment.

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