Abcd is a trapezium with ab parallel to cd. if ac and bd intersect at e and triangle aed is similar to triangle bec prove that ad=bc
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Thank you for asking this question. Here is your answer:
∠1 = ∠2 (these are alternate angles)
∠3 = ∠4 (these are also alternate angles)
and ∠CED = ∠AEB (these are vertically opposite angles)
Δ EDC ≈ ΔEBA
= ED/EB = EC/EA ⇒ ED/EC = EB/EA (equation 1)
We are given this that ΔAED ≈ ΔBEC
ED/EC = EA/EB = AD/BC (this is an equation 2)
So from the equation 1 and equation 2 we will get this:
EB/EA = EA/EB ⇒ (EB)² = (EA)² ⇒ EB = EA
Now we will substitute EB = EA in equation 2, then we will get:
EA/EA = AD/BC ⇒ AD/BC = 1 ⇒ AD = BC
Hence, Proved.
If there is any confusion please leave a comment below.
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