Question

Prove in a triangle ABC, if $AB>AC$ and D is any point on the side BC, then prove that $AB>AD$?

Answer 1

Hint:We start solving the problem by drawing all the given information to get a better view. We use the fact that the angle opposite to the larger side is large to get the relation between angles $\mathrm{\angle }ACB$ and $\mathrm{\angle }CBA$. We then find the relation between angle $\mathrm{\angle }ADB$ and $\mathrm{\angle }ACD$ using the fact that some of the angles in a triangle is ${180}^{\circ }$ and the angle in a straight line is ${180}^{\circ }$. Using this relation we find the relation between the angles $\mathrm{\angle }ADB$ and $\mathrm{\angle }CBA$, which leads us to the desired result.

Complete step by step answer:
According to the problem we have a triangle $ABC$ with sides $AB>AC$ and D is any point on the side BC. We need to prove $AB>AD$.
Let us draw the given information to get a better view.

According to the problem, we have a relation between sides AB and AC as $AB>AC$. We know that the angle opposite to the larger side is large. Applying this fact in $\mathrm{\Delta }ABC$, we get the relation between the angles as $\mathrm{\angle }ACB>\mathrm{\angle }CBA$ ---(1).
We know that the angle in a straight line at a point is ${180}^{\circ }$. We apply this at point D.
We get $$\mathrm{\angle }ADB+\mathrm{\angle }CDA={180}^{\circ }$$.
$$\Rightarrow \mathrm{\angle }ADB={180}^{\circ }-\mathrm{\angle }CDA$$ ---(2).
We know that the sum of the angles in a triangle is ${180}^{\circ }$. We use this in $\mathrm{\Delta }ADC$.
So, we get $$\mathrm{\angle }ACD+\mathrm{\angle }CDA+\mathrm{\angle }DAC={180}^{\circ }$$.
$\Rightarrow \mathrm{\angle }ACD+\mathrm{\angle }DAC={180}^{\circ }-\mathrm{\angle }CDA$.
From equation (2), we get
$\Rightarrow \mathrm{\angle }ACD+\mathrm{\angle }DAC=\mathrm{\angle }ADB$ ---(3).
From the figure, we can see that $\mathrm{\angle }ACD=\mathrm{\angle }ACB$. We use this in equation (3).
$\Rightarrow \mathrm{\angle }ACB+\mathrm{\angle }DAC=\mathrm{\angle }ADB$ ---(4).
From equation (4), we can see that the value of the angle $\mathrm{\angle }ADB$ greater than the angle $\mathrm{\angle }ACB$.
So, we get $\mathrm{\angle }ADB>\mathrm{\angle }ACB$ ---(5).
From equation (1) and (5), we can clearly see that $\mathrm{\angle }ADB>\mathrm{\angle }CBA$ ---(6).
Now, let us consider $\mathrm{\Delta }ADB$ and we have $\mathrm{\angle }ADB>\mathrm{\angle }CBA$. We know that the sides opposite to the larger angle is large. We can see that the side opposite to the angle $\mathrm{\angle }ADB$ is AB and the side opposite to $\mathrm{\angle }CBA$ is AD.
So, we get $AB>AD$.
∴ We have proved $AB>AD$.

Note:
 We should not confuse with the notation of angles present inside the triangle. Whenever we get this type of problem, we should make use of the diagram and the fact that the angle opposite to the larger side is large. We should not assume the angle at point D as ${90}^{\circ }$. We should not consider the line segment AD as the angular bisector as this will not be possible always.