Question

∆ABC is an isosceles triangle in which AB = AC. Side BA is produced to D such
that AD = AB. Show that ∠BCD = 90°.

Answer 1

By theorem,
we know that,
Angle opposite to equal sides are also equal.

Let <ACB = x
then,

<ABC + <ACB + <BAC = 180° (Angle sum property)

x + x + <BAC = 180°

2x = 180 - <BAC

[x = (180 - <BAC)/2] _____ (1)

&

Let <ACD = y
then,

<ACD + <ADC + <CAD = 180° (Angle sum property)

y + y + <CAD = 180°

2y = 180 - <CAD

[y = (180 - <CAD)/2] _____ (2)

Now,
Joining x & y,

$x = \frac{(180 - < bac)}{2} \\ \\ y = \frac{(180 - < cad)}{2} \\ \\ now.joining. \\ \\ x + y = \frac{(180 - < bac)}{2} + \frac{(180 - < cad)}{2} \\ \\ x + y = \frac{180 - < bac + 180 - < cad}{2} \\ \\ x + y = \frac{360 - ( < bac + cad)}{2} \\ \\ by \: figure \: we \: can \: see \: that.. \\ \\ < bac + < cad = 180 \: (linear \: pair) \\ \\ now.putting \: values. \\ \\ we \: have. \\ \\ x + y = \frac{360 - 180}{2} \\ \\ x + y = \frac{180}{2} \\ \\ x + y = 90 \: .......... \: (1) \\ \\ x + y = < bcd \\ \\ we \: can \: replace \: eqn. \: (1) \: with \: < bcd \\ \\ now. \\ \\ ((< bcd = 90))$


Hence proved..

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Answer 2

Hello mate ^_^

__________________________/\_

$\bold\pink{Solution:}$

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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