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SAI International School
CLASS - X
Mathematics
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CHAPTER-6: Triangless-4Lesson Notes-4
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SUBTOPIC : Similar Triangles,BPT & its Conversee
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Pythagoras Theorem
Theorem-6.7
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If a perpendicular is drawn from the vertex of right angle of a right angled
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t t triangle to the hypotenuse, then triangles on both sides of the perpendicular
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a r e s i m i l a r t o t h e w h o l e t r i a n g l e a n d t o e a c h o t h e r
are similar to the whole triangle and to each other..
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Given: A right angled ABC, right angled at B and BD is perpendicular to the
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hypotenuse AC
To prove:(i) ADB ~ ABC (ii) BDC ~ ABC (iii) ADB ~ BDC
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Proof:(i) Consider ADB and ABC, we have,
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A = A (common)
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ADB = ABC (each is 900 )
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Therefore, by AA Criterion of similarity, we have,,
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ADB ~ ABC
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(ii)Similarily BDC ~ ABC by AA Criterion of similarity
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(iii)Now, as ADB ~ ABC and BDC ~ ABC
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So ) ADB ~ BDC
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[If one polygon is similar to another polygon and this second polygon is similar to a
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third polygon, then the first polygon iss similar to the third polygon]
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Theorem-6.8
1
