Question

ΔABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB. Show
that ∠BCD is a right angle.

Answer 1

In $\tt \triangle ABC$, we have

$\tt AB=AC$ [given]

$\tt \angle ACB=\angle ABC...(i)$ [since angles opposite to equal sides are equal]

Now,

$\tt AB=AD$ [given]

$\tt \therefore AD=AC$ [since $\tt AB=AC$]

Thus, in $\tt \triangle ADC$, we have,

$\tt AD=AC$

$\tt \implies \angle ACD =\angle ADC...(ii)$ [since angles opposite to equal sides are equal]

Adding (i) and (ii),

$\tt \angle ACB+\angle ACD=\angle ABC + \angle ADC$ [since $\tt \angle ADC=\angle BDC$ ]

$\tt \implies \angle BCD + \angle BCD= \angle ABC + \angle BDC + \angle BCD$ [Adding $\tt \angle BCD$ on both sides]

$\tt \implies 2\angle BCD =180$ [angle sum property]

$\tt \implies \boxed {\tt{\angle BCD = 90 } }$

Hence, $\tt \angle BCD$ is a right angle.

Answer 2

Answer:

In , we have

[given]

[since angles opposite to equal sides are equal]

Now,

[given]

[since ]

Thus, in , we have,

[since angles opposite to equal sides are equal]

Adding (i) and (ii),

[since  ]

[Adding  on both sides]

[angle sum property]

Hence,  is a right angle.

YUP THATS YOUR ANSWER !

Step-by-step explanation:

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