Page 2 - LN

A

B
C
But [since ABC ~~ PQR] (3)
From (2) and (3), we get,
(4)
From (1) and (4), we get,

(5)
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As ABC ~ PQR, therefore,,
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(6)

(from (5) and (6))
Hence proved.

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*Prove that the areas of two similar triangles are in the ratio of the squares of their
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corresponding altitudes.
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Given: ABC ~ DEF, AL and DM are altitudes to BC and EF respectively..
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To prove:
Proof::

ABC ~ DEF [given]
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As the ratio of the areas of two similar triangles is equal to the ratio of the squares of
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a n y t w o co r r e sp o n d i n g si d e s
any two corresponding sides,,
(i))
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Now, in ABL and DEM,,
[
B = E [given, ABC ~ DEF]
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and ALB = DME [since, both are 90o]

T h e r e f o r e , b y A A c r i t e r i o n o f si m i l a r i t y , w e g e t
Therefore, by AA criterion of similarity, we get,,
ABL ~ DEM

(ii)

2