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`        in isosceles triangle ABC AB=AC. The side BA is produced to D such that BA=BD.Prove that angle BCD=90 degree`
2 years ago

```							In the triangle, We know, AB=AC=AD,Join CD,TAKING,angleB=ACB=ACD=D=xConsider triangle BDC,as a whole, B+ACD+ACB+D=180°ANGLE SUM PROPERTY,ie,x+x,+2x=180°X=45°,We have to prove,
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2 years ago
```							Given in ΔABC, AB = AC⇒ ∠ABC = ∠ACB (Since angles opposite to equal sides are equal)Also given that AD = AB⇒ ∠ADC = ∠ACD (Since angles opposite to equal sides are equal)∴ ∠ABC = ∠ACB = ∠ADC = ∠ACD = x (AB = AC = AD)In ΔBCD, ∠B + ∠C + ∠D = 180°x + 2x + x = 180°4x = 180°x = 45°∠C =  2x = 90°Thus BCD is a right angled triangle
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2 years ago
```							Ok, so Given : AB = AC. , AB = AD             To Prove : BCD = 90.  Proof: In triangle ABC , AB = AC ( Given )......(1)           AB = AD ( Given ) …......(2)From (1) and (2), AC = AD.  Now in Triangle BCD, since AD= AC, ADC = ACD (Angles opposite to equal sides are equal) In triangle ABC, ABC = ACB ( Since AB = AC, angles opposite to equal sides are equal ) Mark ACB and ABC as “x” and ADC and ACD as “y” ( Since both of these pairs of angles are equal ) ….........{3}    Now in triangle BCD, By Angle Sum Property of a traingle , x +x+y+y = 180.  (from {4} ) 2 (x + y) = 180. Therefore x + y 90. But remember that x + y is actually angle BCD ?? Therefore BDC = 90. Hence Proved. Hope this answer helps you.
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2 years ago
```							Given  that in ΔABCAB = ACTherefore we can say that ∠ABC = ∠ACB ( angles opposite to equal sides are equal)Also given that AD = ABThen we can aslo say that ∠ADC = ∠ACD (Since angles opposite to equal sides are equal) ∴ ∠ABC = ∠ACB = ∠ADC = ∠ACD = x (AB = AC = AD)Now in ΔBCD, ∠B + ∠C + ∠D = 180°x + 2x + x = 180°4x = 180°x = 45°∠C =  2x = 90°Thus BCD is a right angled triangle Hence Proved!!
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2 years ago
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