Question

Triangle ABC is an isoscale in which AB=AC . Side BA is produced to D such that AD = AB THEN SHOW THAT ANGLE BCD is a right angle

Answer 1

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

Answer 2

Hello mate ^_^

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$\bold\pink{Solution:}$

AB=AC         (Given)

It means that ∠DBC=∠ACB           (In triangle, angles opposite to equal sides are equal)     

Let ∠DBC=∠ACB=x         .......(1)

AC=AD          (Given)

It means that ∠ACD=∠BDC         (In triangle, angles opposite to equal sides are equal)     

Let ∠ACD=∠BDC=y           ......(2)

In ∆BDC, we have

∠BDC+∠BCD+∠DBC=180°     (Angle sum property of triangle)

⇒∠BDC+∠ACB+∠ACD+∠DBC=180°

Putting (1) and (2) in the above equation, we get

y+x+y+x=180°

⇒2x+2y=180°

⇒2(x+y)=180°

⇒(x+y)=180/2=90°

Therefore, ∠BCD=90°

hope, this will help you.☺

Thank you______❤

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