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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

Lesley
19 Points
4 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,
Shrey Thakkar
42 Points
3 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
C.Vishaal
39 Points
3 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.

Chelsi Kothari
38 Points
3 years ago
Given  that in ΔABC
AB = AC
Therefore we can say that ∠ABC = ∠ACB ( angles opposite to equal sides are equal)
Also given that AD = AB
Then 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!!