Given: The correspondence ABC ↔ DEF is similarity in ∆ABC and ∆DEF.
is an altitude of ∆ABC and is an altitude of ∆DEF.
To Prove: AB × DN = AM × DE
Proof: Since, the correspondence ABC ↔ DEF is a similarity.
By definition, for a given correspondence between the vertices of two triangles, if the corresponding angles of the triangles are congruent and the lengths of the corresponding sides are in proportion, then the given correspondence is a similarity between two triangles.
So, we can write ∠B ≅ ∠E. …(i)
In ∆ABM and ∆DEN,
From result (i),
∠B ≅ ∠E
Also, ∠M ≅ ∠N [since, they are right angles of ∆ABM and ∆DEN respectively]
⇒ By AA-corollary, the correspondence ABM ↔ DEN is a similarity in ∆ABC and ∆DEF]
Then, again by definition of similarity of correspondences in triangles we can say that,
By cross-multiplication, we get
⇒ AB × DN = AM × DE
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