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NCERT/CBSE MATHEMATICS CLASS 10 textbook(OPTIONAL EXERCISE) http://www.TutorBreeze.com Contact for Online Tutoring in Physics, Math, Chemistry, English Answers to NCERT/CBSE MATHEMATICS Class 10(Class XI)textbook OPTIONAL EXERCISE CHAPTER SIX SIMILAR TRIANGLES EXERCISES

2. In the given Figure, D is a point on hypotenuse AC of ∆ ABC, DM ⊥ BC and DN ⊥ AB. Prove that : 2 2 (i) DM = DN . MC (ii) DN = DM . AN .

Given: D is a point on hypotenuse AC of ∆ ABC, DM ⊥ BC and DN ⊥ AB. 2 2 To prove(i) DM = DN . MC (ii) DN = DM . AN

Proof: DM ⊥ BC and DN ⊥ AB. ⇒BMDN is a rectangle ©TutorBreeze.com Please do not copy the answer given here Write to us for help in understanding the solution

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⇒BM = ND (i) In triangle BMD ∠ M+ ∠1+∠2= 180゚[ASP ] ⇒∠1+∠2 = 90゚ In the same way, ∠3+∠4=90゚ Also, ∠3+∠2=90゚ ⇒∠1+∠2

=∠3+∠2

⇒∠1 =∠3 Similarly, ∠3+∠4=∠3+∠2 ⇒ ∠4=∠2 Now in ∆BMD and ∆DMC ∠1

=∠3 and ∠4=∠2 ∴∆BMD ∼ ∆DMC[AA Criterion]

BM MD = DM MC ButBM = ND ND MD ⇒ = DM MC ⇒ DM 2 = DN × MC

(ii)

Proceeding in the same way as in (i),We can similarly prove that

∴∆BND ∼ ∆DNA[AA Criterion]

BN ND = DN NA ButDM = BN DM ND ⇒ = DN NA 2 ⇒ DN = DM × AM

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