Question

Triangle ABC is an isoceles triangle in which AB=AC .side BA is produced to D such that AD = AB . Show that angle BCD is right angle

Answer 1

Construction :- Join DC.

Proof :- We have , AB = AC (given)

also, AB = AD

→AD =AC →<ACD = <ADC

.....................(1){Angles opposite to equal sides of ∆}

Similarly,

< ABC = <ACB........(2)

From (1) and (2) we get ,

<C = <D +<B .....(3)

Now,

<D +< B +<C = 180(<s sum property of ∆)

<C +<C = 180{using 3}

→ C = 90

→BCD is a right ∆.

Answer 2

Ok the solution is here

Figure is given in the attachment.

IN ∆ABC

As AB=AC

ANGLE ABC=ANGLE ACB...EQ1

(ANGLES OPPOSITE TO EQUAL SIDES ARE EQUAL)

AND IN ∆

AS AD=AB

AD=AC(:::AB=AC)

ANGLE ACD = ANGLE ADC ....EQ2

(ANGLES OPPOSITE TO EQUAL SIDES ARE EQUAL)

AS WE KNOW SUM OF ALL ANGLES OF A TRIANGLES=180°

SO IN ∆BCD

ANGLE ABC+ANGLE ACD+ANGLE ACD+ANGLE CDA=180°

FROM EQ1 AND EQ2

WE GET

2ANGLE ACD +2ANGLE ACB=180°

2(ANGLE ACD+ANGLE ACB)=180°

ANGLE BCD=180°/2

ANGLE BCD=90°

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