Question

AD bisects angle $$A$$ of triangle$$ABC$$, where $$D$$ lies on $$BC$$ and angle $$C$$ is greater than angle $$B$$, then show that the angle $$ADB$$ is greater than angle $$ADC.$$

Answer 1

Hint: At first we will try to find out the value of the angles of the triangle $$ADB$$ and triangle $$ADC$$in terms of angles$$A,B\mathrm{\&}C$$. Then we will find the relation between the required angles.

Complete step-by-step answer:
It is given that, $$AD$$ bisects angle $$A$$ of triangle$$ABC$$, where $$D$$lies on $$BC$$ and angle $$C$$ is greater than angle$$B$$.

We know that sum of the all the angles of triangle is $${180}^{\circ }.$$
Since, $$AD$$ bisects angle $$A$$ of triangle$$ABC$$,
We can write $$\mathrm{\angle }DAB={\displaystyle \frac{1}{2}}\mathrm{\angle }A$$
So, now let us consider the triangle $$ADB$$ in that since we have,
Sum of all the angles is $${180}^{\circ }.$$
That is the sum of the angles in the triangle$$ABC$$ ,
$$\mathrm{\angle }ABD+\mathrm{\angle }ADB+\mathrm{\angle }BAD={180}^{\circ }.$$
We know that$$\mathrm{\angle }BAD={\displaystyle \frac{1}{2}}\mathrm{\angle }A$$ and $$\mathrm{\angle }ABD=\mathrm{\angle }B$$on substituting the known values we get,
$$\mathrm{\angle }ADB={180}^{\circ }-\mathrm{\angle }B-{\displaystyle \frac{1}{2}}\mathrm{\angle }A$$
Similarly, from the triangle $$ADC$$ using the fact that Sum of all the angles is $${180}^{\circ }.$$we have,
$$\mathrm{\angle }ACD+\mathrm{\angle }ADC+\mathrm{\angle }CAD={180}^{\circ }.$$
Here we know that $$\mathrm{\angle }CAD={\displaystyle \frac{1}{2}}\mathrm{\angle }A$$ and $$\mathrm{\angle }ACD=\mathrm{\angle }C$$ on substituting the values in the above equation we get,
$$\mathrm{\angle }ADC={180}^{\circ }-\mathrm{\angle }C-{\displaystyle \frac{1}{2}}\mathrm{\angle }A$$
It is also given that,$$\mathrm{\angle }C>\mathrm{\angle }B$$
Let us multiply by $$-1$$on both sides of the inequality, we get,
$$-\mathrm{\angle }C<-\mathrm{\angle }B$$
Now let us add $${180}^{\circ }$$ in both sides of the inequality, then we have,
$${180}^{\circ }-\mathrm{\angle }C<{180}^{\circ }-\mathrm{\angle }B$$
Also let us subtract $${\displaystyle \frac{1}{2}}\mathrm{\angle }A$$ in both sides of the above inequality then we have,
$${180}^{\circ }-\mathrm{\angle }C-{\displaystyle \frac{1}{2}}\mathrm{\angle }A<{180}^{\circ }-\mathrm{\angle }B-{\displaystyle \frac{1}{2}}\mathrm{\angle }A$$
We know that from equation (1) and (2) we have $$\mathrm{\angle }ADB={180}^{\circ }-\mathrm{\angle }B-{\displaystyle \frac{1}{2}}\mathrm{\angle }A$$ and$$\mathrm{\angle }ADC={180}^{\circ }-\mathrm{\angle }C-{\displaystyle \frac{1}{2}}\mathrm{\angle }A$$
On substituting the values in the inequality we get,
$$\mathrm{\angle }ADB>\mathrm{\angle }ADC$$.

Hence, we have shown that the angle $$ADB$$ is greater than angle $$ADC.$$

Note:
Let us consider, $$A>B$$ then by multiplying $$-1$$ with both sides we will have the relation as,$$-B>-A$$ that is the relation greater than is changed to less than and vice versa on multiplying the inequality by $$-1$$