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Abc is a Triangle in Which Be and Cf Are, Respectively, the Perpendiculars to the Sides Ac And Ab. If Be = Cf, Prove that δAbc is Isosceles - Mathematics

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प्रश्न

ABC is a triangle in which BE and CF are, respectively, the perpendiculars to the sides AC and AB. If BE = CF, prove that ΔABC is isosceles  

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उत्तर

Given that ABC is a triangle in which BE and CF are perpendicular to the sides AC and AB respectively such that BE = CF   

We have to prove that ΔABC is isosceles
Now, consider ΔBCF and ΔCBE, 

We have 

∠BFC=CEB=90°                 [Given] 

BC=CB                             [Common side] 

And CF=BE                        [Given]  

So, by RHS congruence criterion, we have ΔBFC≅CEB 

Now, 

∠FBC=∠EBC   [∵ Incongruent triangles corresponding parts                                                   are equal] 

⇒ ∠ABC=∠ACB 

⇒ AC=AB    [ ∵Opposite sides to equal angles are equal in a                                triangle]

∴ ΔABC is isosceles 

 

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पाठ 12: Congruent Triangles - Exercise 12.5 [पृष्ठ ६१]

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आरडी शर्मा Mathematics [English] Class 9
पाठ 12 Congruent Triangles
Exercise 12.5 | Q 2 | पृष्ठ ६१

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ABC is an isosceles triangle with AB = AC and D is a point on BC such that AD ⊥ BC (Figure). To prove that ∠BAD = ∠CAD, a student proceeded as follows:


In ∆ABD and ∆ACD,

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and ∠ADB = ∠ADC

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[Hint: Recall how ∠B = ∠C is proved when AB = AC].


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