Question

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question number 6

Answer 1

In ∆ABC & ∆DAC
AB=AC(GIVEN)
AD=AB(GIVEN)
AC=AC(GIVEN)
∆ABC=~ ∆DAC(by SSS rule)
In ∆BCD
<B+<D+<C=180°(BY <sum property of ∆)
x+y+x+y=180°
2(x+y)=180°/2=90°
<C=90°
Thus it is proved <BCD is right <'d triangle.........

Answer 2

Hello Dear 
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Question 6 : ΔABC is an isosceles triangle in which AB = AC . Side BA is produced  to D such that AD = AB (see fig . 7.34) . SHow that ∠BCD is a right angle . 
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Answer :  AB = AC = AD [Given] 

∠BAC + ∠ABC + ∠ACB  = 180 ------1

∠ACD + ∠CAD + ∠ADC = 180 ------2

∠ACB + ∠ACD + 180 + 90 = 360 [Adding 1 and 2 ]

∠ACB + ∠ACD = 360 - 180 - 90 

∠ACB + ∠ACD = 90 

∠DCB = 90° 
Hence Proved 
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Question 7 : ABC is a right angled triangle in which ∠A = 90 and AB = AC
Find ∠B and ∠C 
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