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# In the figure, it is given that $TS = TR$, $\angle 1 = 2\angle 2,\angle 4 = 2\angle 3$. Prove that $\Delta RBT = \Delta SAT$.

Last updated date: 19th Jun 2024
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Hint:
We have given congruence about two angles and equality about one side of a triangle. So we will make the adjustments in the given data so that we will use one of the tests for congruence by using given data and then we will prove the congruence of the triangles.

Complete step by step solution:
First of all, we will start by noting down the given data.
It is given that $TS = TR$, $\angle 1 = 2\angle 2$ and $\angle 4 = 2\angle 3$ .
Now since it is given that $TS = TR$, the given triangle $\Delta TRS$ is an isosceles triangle.
Therefore, the base angles are also equal.

That means we can write the following:
$\angle TSR = \angle TRS$ … (1)
Now the angles $\angle 1$ and $\angle 4$ are vertically opposite angles and therefore, are equal.
Now it is given that $\angle 1 = 2\angle 2$ and $\angle 4 = 2\angle 3$ .

But since we just established that $\angle 4 = \angle 1$ we can write,
$2\angle 2 = 2\angle 3$
Dividing both sides by $2$ we get,
$\angle 2 = \angle 3$ … (2)
Subtract equation (2) from equation (1) and write,
$\angle TSR - \angle 3 = \angle TRS - \angle 2$
From the figure we observe that $\angle TRS - \angle 2 = \angle TRB$ and $\angle TSR - \angle 3 = \angle TSA$ .

Now in $\Delta RBT$ and $\Delta SAT$ we observe that $\angle STA = \angle RTB$ as it is a common angle.
Also, we just showed that $\angle TSA = \angle TRB$ .
We have already given that $RT = ST$ .

Therefore, by using the $ASA$ test for congruence, we conclude that $\Delta RBT = \Delta SAT$.

Note:
Here we had already given the congruent side and other ratios were about the angles. So, it is a hint indicating that we have to use a test for congruence which kind of involves only one side and two angles. Therefore, we will make the adjustments in the given equations accordingly.