Question

Triangle abc is an isosces triangle in which ab=ac.Side ba is produced to d such that ad=ab.Shiw that angle bcd is a right angle

Answer 1

Answer:

ab=ac

ad=ab

ab=ac=ad

=><b=<acb=<acd=<d;<acd+<acb=<c

let it be x,

<b+<c+<d=180

x+2x+x=180

4x=180

x=45

<bcd=<c=2x=90°

Answer 2

GIVEN:-

ABC is isosceles ∆

AB=AC

AD=AB

TO PROVE:-

$\large\sf{\angle{BCD}=90°}$

PROOF:-

$\large\sf{In\:∆ABC,}$

$\large\sf{AB=AC(given)}$

$\large\sf{\angle{ABC}=\angle{ACB}(angle\:opp.\:to\:equal\:sides)}$.......(1)

⠀

$\large\sf{In\:∆ACD,}$

$\large\sf{AD=AC(AD=AB\:and\:AB=AC)}$

$\therefore$$\large\sf{\angle{ACD}=\angle{ADC}(angle\:opp.\:to\:equal\:side)}$......(2)

$\huge\sf\red{We\:know\:that,}$

$\large\sf{\angle{DBC}+\angle{BCD}+\angle{CDB}=180°}$

$\large\sf{\angle{BCD}+\angle{ABC}+\angle{CDA}=180°}$

$\large\sf{\angle{BCD}+\angle{ACB}+\angle{ACD}=180°(from(i)and(ii)}$

$\large\sf{\angle{BCD}+\angle{BCD}=180°}$

$\large\sf{2\angle{BCD}=180°}$

$\large\sf{\angle{BCD}=\frac{180°}{2}}$

$\large\sf{\angle{BCD}=90°}$