Question

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Answer 1

Good question.

Let me do the following.

The respective figure edited is attached with this.

Join ED, AC and BD.

Here, C, D, E are collinear. So ED passes through C.

ED, the radius of outer circle, is the diameter of the inner circle.

∴ EC = DC = 1/2 × ED

Here it seems that the radius of the inner circle is 1/2 times that of the outer circle.

AC and BD are also radii of the inner and outer circles respectively.

∴ AC = 1/2 BD

Okay, consider ΔBDE.

BD = ED (radius of the same circle)

∴ ∠B = ∠E     →     (1)

Consider ΔACE.

AC = EC (radius of the same circle)

∴ ∠E = ∠A     →     (2)

From (1) and (2),

∠A = ∠B = ∠E

Consider both triangle ACE and BDE.

∠E = ∠E (common)

∠A = ∠B (Found earlier)

∴ ΔACE ~ ΔBDE

∴ ∠C = ∠D

∴ AC ║ BD

We know that, if a line is drawn parallel to one side of a triangle from the midpoint of another side, then it reaches the midpoint of the third side, and the length of this line segment is 1/2 the length of the side parallel to it.

We found that C is the midpoint of ED, AC ║ BD and AC = 1/2 BD.

∴ According to the above theorem, A is the midpoint of EB.

∴ EA = AB.

∴ EA ≅ AB.

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Hope this may be helpful.

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Thank you. Have a nice day. :-)

 

#adithyasajeevan

Answer 2

Construct ED as diameter and draw AD
Therefore angle EAD is angle in semicircle therefore EAD is 90 degree
DA is perpendicular to EB.
EA=AB