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Maths-

General

Easy

Question

D and E are points on the base BC of $straight triangle space ABC$ such that BD = CE. If AD = AE, prove that $straight angle ABE equals straight angle ACD$ .

hint Hint:

prove the given condition by using similarity of triangles

The correct answer is: Hence proved

In $straight capital delta$ ACD and $straight capital delta$ ABE
AD = AE (given)
BD = EC (given)
Now adding DE on both side to above conditions ,
BD + DE = DE + EC
BE = CD
∠ADC = ∠AEB (as AD = AE)
By SAS rule , $straight capital delta$ ABE ≅ $straight capital delta$ ACD
Then , by congruence ∠ABE = ∠ACD
Hence proved

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