Question
In $\triangle\text{ABC},$ side $AB$ is produced to $D$ so that $\text{BD}=\text{BC}.$ if $\angle\text{B}=60^\circ$ and $\angle\text{A}=70^\circ.$ Prove that: $(i) AD > CD (ii) AD > AC$
✓
Answer
Given that, in $\triangle\text{ABC},$ side $AB$ is produced to $D$ so that $\text{BD}=\text{BC}.$
$\angle\text{B}=60^\circ,$ and $\angle\text{A}=70^\circ$
To prove, $AD > CD AD > AC$ First join $C$ and $D$
We know that, Sum of angles in a triangle $=180^\circ $
$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
$70^\circ+60^\circ+\angle\text{C}=180^\circ$
$\angle\text{C}=180^\circ-(130^\circ)=50^\circ$
$\angle\text{C}=50^\circ$
$\angle\text{ACB}=50^\circ\ ...(\text{i)}$ And also in $\triangle\text{BDC}$
$\angle\text{DBC}=180^\circ-\angle\text{ABC}$ [$ABD$ is a straight angle]
$180^\circ-60^\circ=120^\circ$ and also $\text{BD}=\text{BC}$ [given]
$\angle\text{BCD}=\angle\text{BDC}$ [Angles opposite to equal sides are equal]
Now, $\angle\text{DBC}+\angle\text{BCD}+\angle\text{BDC}=180^\circ$ [Sum of angles in a triangle $=180^\circ $] $\Rightarrow120^\circ+\angle\text{BCD}+\angle\text{BCD}=180^\circ$
$\Rightarrow120^\circ+2\angle\text{BCD}=180^\circ$
$\Rightarrow2\angle\text{BCD}=180^\circ-120^\circ=60^\circ$
$\Rightarrow\angle\text{BCD}=30^\circ$
$\Rightarrow\angle\text{BCD}=\angle\text{BDC}=30^\circ\ ....(\text{ii)}$
Now, consider $\triangle\text{ADC}.$
$\angle\text{BAC}\Rightarrow\angle\text{DAC}=70^\circ$ [given] $\angle\text{BDC}$
$\Rightarrow\angle\text{ADC}=30^\circ$ [From $(ii)]$
$\angle\text{ACD}=\angle\text{ACB}+\angle\text{BCD}$ $= 50^\circ + 30^\circ$ [From $(i)$ and $(ii)] = 80^\circ$
Now, $\angle\text{ADC}<\angle\text{DAC}<\angle\text{ACD}$
$\text{AC}<\text{DC}<\text{AD}$ [Side opposite to greater angle is longer and smaller angle is smaller]
$\text{AD}>\text{CD}$ and $\text{AD}>\text{AC}$
Hence proved Or,
We have, $\angle\text{ACD}>\angle\text{DAC}$ and $\angle\text{ACD}>\angle\text{ADC}$
$\text{AD}>\text{DC}$ and $\text{AD}>\text{AC}$
[Side opposite to greater angle is longer and smaller angle is smaller]
$\angle\text{B}=60^\circ,$ and $\angle\text{A}=70^\circ$
To prove, $AD > CD AD > AC$ First join $C$ and $D$
We know that, Sum of angles in a triangle $=180^\circ $
$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
$70^\circ+60^\circ+\angle\text{C}=180^\circ$
$\angle\text{C}=180^\circ-(130^\circ)=50^\circ$
$\angle\text{C}=50^\circ$
$\angle\text{ACB}=50^\circ\ ...(\text{i)}$ And also in $\triangle\text{BDC}$
$\angle\text{DBC}=180^\circ-\angle\text{ABC}$ [$ABD$ is a straight angle]
$180^\circ-60^\circ=120^\circ$ and also $\text{BD}=\text{BC}$ [given]
$\angle\text{BCD}=\angle\text{BDC}$ [Angles opposite to equal sides are equal]
Now, $\angle\text{DBC}+\angle\text{BCD}+\angle\text{BDC}=180^\circ$ [Sum of angles in a triangle $=180^\circ $] $\Rightarrow120^\circ+\angle\text{BCD}+\angle\text{BCD}=180^\circ$
$\Rightarrow120^\circ+2\angle\text{BCD}=180^\circ$
$\Rightarrow2\angle\text{BCD}=180^\circ-120^\circ=60^\circ$
$\Rightarrow\angle\text{BCD}=30^\circ$
$\Rightarrow\angle\text{BCD}=\angle\text{BDC}=30^\circ\ ....(\text{ii)}$
Now, consider $\triangle\text{ADC}.$
$\angle\text{BAC}\Rightarrow\angle\text{DAC}=70^\circ$ [given] $\angle\text{BDC}$
$\Rightarrow\angle\text{ADC}=30^\circ$ [From $(ii)]$
$\angle\text{ACD}=\angle\text{ACB}+\angle\text{BCD}$ $= 50^\circ + 30^\circ$ [From $(i)$ and $(ii)] = 80^\circ$
Now, $\angle\text{ADC}<\angle\text{DAC}<\angle\text{ACD}$
$\text{AC}<\text{DC}<\text{AD}$ [Side opposite to greater angle is longer and smaller angle is smaller]
$\text{AD}>\text{CD}$ and $\text{AD}>\text{AC}$
Hence proved Or,
We have, $\angle\text{ACD}>\angle\text{DAC}$ and $\angle\text{ACD}>\angle\text{ADC}$
$\text{AD}>\text{DC}$ and $\text{AD}>\text{AC}$
[Side opposite to greater angle is longer and smaller angle is smaller]
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