Video: CBSE Class X • Pack 1 • 2018 • Question 30A

CBSE Class X • Pack 1 • 2018 • Question 30A

08:43

Video Transcript

In an equilateral triangle 𝐴𝐵𝐶, there is a point 𝐷 on the side 𝐵𝐶 such that 𝐵𝐷 is equal to one-third 𝐵𝐶. Prove that nine 𝐴𝐷 squared is equal to seven 𝐴𝐵 squared.

Here, we have the triangle 𝐴𝐵𝐶. And since it’s an equilateral triangle, this means that the lengths of each of the sides are the same. And this also means that all of the interior angles of this triangle are 60 degrees.

Now that we have drawn 𝐴𝐵𝐶, let’s locate the point 𝐷. The piece of information that we’re told about the point 𝐷 is that 𝐵𝐷 is equal to one-third 𝐵𝐶. We are also told that the point 𝐷 is on the side 𝐵𝐶. 𝐵𝐷 is one-third of 𝐵𝐶. So this tells us that the point 𝐷 is one-third of the way from 𝐵 to 𝐶.

And so we can draw our point 𝐷 on the triangle here. And we can label the side 𝐵𝐷 as the length one-third 𝐵𝐶. We can also find the length 𝐷𝐶 since 𝐵𝐶 is equal to 𝐵𝐷 plus 𝐷𝐶. And we know that 𝐵𝐷 is equal to one-third 𝐵𝐶. And so by subtracting one-third 𝐵𝐶 from both sides of this equation, we obtain that 𝐷𝐶 is equal to two-thirds 𝐵𝐶.

Now that we have found the lengths 𝐵𝐷 and 𝐷𝐶, we can start to look at what the question has asked us. We have been asked to show that nine 𝐴𝐷 squared is equal to seven 𝐴𝐵 squared. We can label the sides 𝐴𝐵 and 𝐴𝐷 on our diagram. And we can see that 𝐴𝐵 is one of the sides of the equilateral triangle.

And one of the properties of equilateral triangles, as discussed earlier, is that all of the side lengths are the same. And so we have that 𝐴𝐵 is equal to 𝐵𝐶 is equal to 𝐴𝐶. And since 𝐴𝐵 is equal to 𝐵𝐶, we can rewrite the length 𝐵𝐷 and 𝐷𝐶 in terms of 𝐴𝐵. This will come in handy later on since the thing we’re trying to prove is only in terms of 𝐴𝐷 and 𝐴𝐵. So by converting as many things as we can into 𝐴𝐵 or 𝐴𝐷, we’ll make it easier later on.

Now that we have these lengths in terms of 𝐴𝐵, we need to try to find a way of getting 𝐴𝐷 in terms of 𝐴𝐵. In order to do this, we can draw some right-angled triangles to help us out. Let’s draw a line vertically down from the point 𝐷 to the line 𝐴𝐶. And we can call the point where it intersects the line 𝐴𝐶 the point 𝑋.

And now since this is a vertical line hitting a horizontal line, this means that the two lines are perpendicular. And so we know that the angle 𝐷𝑋𝐶 and the angle 𝐷𝑋𝐴 will both be right angles. Now, we can see that we have formed two more triangles. Let’s start by looking at triangle 𝐷𝐶𝑋.

Here, we have the triangle 𝐷𝐶𝑋. We’ve labelled the side 𝐷𝐶 with a length two-thirds 𝐴𝐵, which we found earlier. And we have the angle 𝐷𝐶𝑋 is 60 degrees and the angle 𝐶𝑋𝐷 is 90 degrees.

Now, in order to find 𝐴𝐷 in terms of 𝐴𝐵 and we look at our original diagram, we can see that by using the dashed orange triangle — so that’s the triangle 𝐴𝐷𝑋, which is in fact a right-angled triangle — if we know the length 𝐷𝑋 and 𝐴𝑋, then we can use Pythagoras’s in order to find the length 𝐴𝐷. And so therefore, if we find the length 𝐷𝑋 and 𝐶𝑋, we will be able to find the length required to find 𝐴𝐷.

Here, we have a right-angled triangle. And we know one of the side lengths and one of the angles other than the right angle. We are able to find the length 𝐷𝑋 and 𝐶𝑋 using SOHCAHTOA. Let’s start by labelling the hypotenuse, the opposite, and the adjacent sides on our triangle.

Let’s start by finding the length of the side 𝐷𝑋. Now, 𝐷𝑋 is the opposite side to the angle which we know and we also have the hypotenuse. And the trig ratio which uses the opposite and the hypotenuse is SOH. And this tells us that the opposite is equal to the hypotenuse timesed by sin of the angle. And so we have that the opposite 𝐷𝑋 is equal to the hypotenuse which is two-thirds 𝐴𝐵 timesed by sin of the angle, which is sin of 60 degrees.

Now, we know that sin of 60 is equal to root three over two. And so we can simplify this equation to obtain that 𝐷𝑋 is equal to root three over three 𝐴𝐵. And so we can label this on our diagram. Next, let’s find the length 𝐶𝑋.

Now, 𝐶𝑋 is the adjacent side to the angle. And we also know the opposite side and the hypotenuse. Now, we could use either CAH or TOA here. However, since the hypotenuse did not involve a square root, it might be easier to use CAH. We have that the adjacent side is equal to the hypotenuse timesed by cos of the angle. So that means that 𝐶𝑋 — the adjacent — is equal to the hypotenuse, which is two-thirds 𝐴𝐵, timesed by cos of the angle — so that’s cos of 60.

And now, we know that cos of 60 is equal to one-half. So we can substitute this in here. And we can simplify this equation to obtain that 𝐶𝑋 is equal to one-third 𝐴𝐵. And so we can label this on our triangle. Now we’ve labelled these lengths, back onto our original diagram. And we can see that we can use the length of 𝐶𝑋 in order to find the length of 𝐴𝑋 since 𝐴𝐶 is equal to 𝐴𝑋 plus 𝐶𝑋.

And since this is an equilateral triangle, we have that 𝐴𝐵 is equal to 𝐴𝐶. And so we can substitute this in here. And we have also already found the length of 𝐶𝑋 to be one-third 𝐴𝐵. And so this can also be substituted in. And now, we simply subtract one-third 𝐴𝐵 from both sides of this equation. And we obtain that 𝐴𝑋 is equal to two-thirds 𝐴𝐵.

Now that we have found the lengths of 𝐴𝑋 and 𝐷𝑋, we are ready to consider the triangle 𝐴𝐷𝑋. Here is the triangle 𝐴𝐷𝑋. And we know the side length of 𝐷𝑋 is root three over three 𝐴𝐵 and the side length of side 𝐴𝑋 is two-thirds 𝐴𝐵. And we also know that the angle 𝐴𝑋𝐷 is a right angle. Since this is a right-angled triangle where we know two of the side lengths, we can use Pythagoras’s here.

What this tells us is that the sum of the square of the two shorter lengths is equal to the square of the longer length. In our case, this means that 𝐴𝐷 squared is equal to 𝐷𝑋 squared plus 𝐴𝑋 squared. And now, we just need to substitute in the side lengths for 𝐷𝑋 and 𝐴𝑋. And we get that 𝐴𝐷 squared is equal to root three over three 𝐴𝐵 all squared plus two-thirds 𝐴𝐵 all squared.

Now, it’s important to remember that we’re squaring the whole side length and not just the 𝐴𝐵s. And so we must put brackets around each of these terms when squaring them. And so squaring each part inside the bracket, we obtain root three squared over three squared times 𝐴𝐵 squared plus two squared over three squared times 𝐴𝐵 squared.

And squaring all of the numbers here, we obtain three-ninths 𝐴𝐵 squared plus four-ninths 𝐴𝐵 squared. And now, we can factorize out the 𝐴𝐵 squared to obtain three-ninths plus four-ninths 𝐴𝐵 squared. And so simplifying this, we obtain that 𝐴𝐷 squared is equal to seven-ninths 𝐴𝐵 squared.

And so the final step in this proof is simply to multiply both sides by nine. And we obtain that nine lots of 𝐴𝐷 squared is equal to seven lots of 𝐴𝐵 squared, which completes the proof here.

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