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Question
In the given figure below, `(AD)/(AE) = (AC)/(BD)` and ∠1 = ∠2, Show that ΔBAE ∼ ΔCAD.
Solution
In ΔABC,
∠1 = ∠2
∴ AB = BD ......(i)
Given, `(AD)/(AE) = (AC)/(BD)`
Using equation (i), we get
`(AD)/(AE) = (AC)/(AB)` ......(ii)
In ΔBAE and ΔCAD, by equation (ii),
`(AC)/(AB) = (AD)/(AE)`
∠A = ∠A ......(Common)
∴ ΔBAE ∼ ΔCAD ......[By SAS similarity criterion]
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In the given figure, ΔLMN is similar to ΔPQR. To find the measure of ∠N, complete the following activity.
Given: ΔLMN ∼ ΔPQR
Since ΔLMN ∼ ΔPQR, therefore, corresponding angles are equal.
So, ∠L ≅ `square`
⇒ ∠L = `square`
We know, the sum of angles of a triangle = `square`
∴ ∠L + ∠M + ∠N = `square`
Substituting the values of ∠L and ∠M in equation (i),
`square` + `square` + ∠N = `square`
∠N + `square` = `square`
∠N = `square` – `square`
∠N = `square`
Hence, the measure of ∠N is `square`.
