In a triangle ABC, AB=BC and <ACB=50 degrees. D is a point on AC such that AD=BD. E is a point on BD such that BE=CD. Find the size of <EAD

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In the triangle ABC (refer to the attached image), AB=BC and <ACB=50 degrees . So, angleBAC=50^o.

Again, in the isosceles `Delta ADB` , <ABD=`50^o` , <ADB=180-(50+50)=`80^o`

Considering `Delta ABE` and `Delta BDC` :

DC=BE

AB=BC

And containing <ABE(i.e.<ABD)=<DCB (i.e. <ACB)

Hence, `Delta ABE ~= Delta BDC`

Therefore, <BAE=<CBD= 180-(100+50)=`30^o`

Hence, <EAD=50-30`=20^o` .

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