Question

The correspondence ABC ⇔DEF is similarity in ΔABC andΔDEF. AM is an altitude of AABC and DN is an altitude of ADEF. Prove that AB X DN = AM X DE.

Answer 1

To prove : AB × DN = AM × DE

proof :- The correspondence ABC ⇔DEF is similarity in ΔABC andΔDEF.
AM is an altitude of ∆ABC and DN is an altitude of ∆DEF as shown in figure.

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 of equation (i),
∠B ≅ ∠E
Also, ∠M ≅ ∠N [ 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,
$\frac{AB}{DE}=\frac{AM}{DN}$
By cross-multiplication, we get
⇒ AB × DN = AM × DE
Hence, proved.

Answer 2

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