Question

triangle abc is an isoscles triengle in which ab equal to ac side ba is produced to d such that ad is equal to ab show that angle bcd is a right angle

Answer 1

In ΔABC, we know that AB=AC.

∴ ∠ABC=∠ACB

Let ∠ABC=Ф,

∴∠ACB=Ф (∵∠ABC=∠ACB)

Now, since AB=AD

then, AD=AC (∵AB=AC)

∴ In ΔADC,

∠ADC=∠ACD

Now, let ∠ADC=α,

∴∠ACD=α (∵ ∠ADC=∠ACD)

So, now angles of ΔBCD are,

∠DBC=Ф

∠DCB=Ф+α (∵ ∠DCB=∠ACD+∠ACB=Ф+α)

∠CDB=α

So according to a property of any Δ,

∠DBC+∠DCB+∠CDB=180°

or, Ф+(Ф+α)+α=180°

or, 2Ф+2α=180°

or, 2(Ф+α)=180°

or, Ф+α=180°/2=90°

∴∠DCB=90°  (∵ ∠DCB=Ф+α)

∴ΔBCD is a right angle Δ with ∠DCB=90°

Here is a photo of the figure for your help,

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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