Chapter 18.5 - Perpendicular from a Vertex to opposite side

In the previous section we saw the Triangle division using parallel lines. In this section, we will see more details.

1. In fig.18.29(a) below, we have a small triangle PQR. We are going to draw lines through each vertices of this triangle. Each of these new lines must be parallel to the opposite side.

Fig.18.29

• First, we draw a line through the vertex R, in such a way that, it is parallel to the side PQ. It is shown in fig.b

• Next, we draw a similar line through the vertex Q, in such a way that, it is parallel to the side PR. This is shown in fig.c

• Next, we draw a similar line through the vertex P, in such a way that, it is parallel to the side RQ. This is shown in fig.d

• Finally, these 3 new lines meet at A, B and C, to form a larger triangle ABC. This is shown in fig.e

■ What we did above was this:

• We drew a line through each vertex

• This line is parallel to the opposite side

2. So by theorem 18.5 that we saw in the previous section, each side of the smaller triangle is half of the corresponding side of the larger triangle. We can write:

3. PQ = ^AC⁄₂ ; QR = ^AB⁄₂ ; PR = ^BC⁄₂

4. Also we have: P is the midpoint of AC; Q is the midpoint of BC; R is the midpoint of AC

5. Now let us look at the fig.e more closely. It is shown in fig.18.30(a) below:

Fig.18.30

6. Consider the inner ΔPQR. Drop a perpendicular from the vertex R, onto the opposite side PQ. It is shown in green colour in fig.b

■ What is the speciality of this perpendicular green line?

• We drew it perpendicular to PQ. So it will be perpendicular to AC also (∵ AC is parallel to PQ)

• Now, from (4), R is the midpoint of AC. So, we have a green line perpendicular to AC, and it is passing through the midpoint of AC

• What do we call such a line?

Ans: It is the perpendicular bisector of AC

7. Now take the vertex P. Drop a perpendicular from the vertex P, onto the opposite side RQ. It is shown in green colour in fig.18.30(c) below:

Fig.18.30

[Note that, in this particular case, the foot of the perpendicular falls out side ΔPQR. So we have to extend side RQ with a dashed red line. Any way, this will not alter the result that we are seeking]

■ What is the speciality of this perpendicular green line?

• We drew it perpendicular to RQ. So it will be perpendicular to AB also (since AB is parallel to RQ)

• Now, from (4), P is the midpoint of AB. So, we have a green line perpendicular to AB, and it is passing through the midpoint of AB

• What do we call such a line?

Ans: It is the perpendicular bisector of AB

8. Now take the vertex Q. Drop a perpendicular from the vertex Q, onto the opposite side PR. It is shown in green colour in fig.18.29(d) above.

[Note that, here also, the foot of the perpendicular falls out side ΔPQR. So we have to extend side PR with a dashed red line. As mentioned earlier, this will not alter the result that we are seeking]

■ What is the speciality of this perpendicular green line?

• We drew it perpendicular to PR. So it will be perpendicular to BC also (since BC is parallel to PR)

• Now, from (4), Q is the midpoint of BC. So, we have a green line perpendicular to BC, and it is passing through the midpoint of BC

• What do we call such a line?

Ans: It is the perpendicular bisector of BC

■ So we have a very interesting situation here:

• Our task was simply to drop perpendicular from the vertices of the inner triangle

• When we completed that task, we got the perpendicular bisector of the sides of the outer triangle as a bonus

9. So the green lines in fig.a serves two purposes:

• They are the perpendiculars from the vertices to the opposite sides of the inner triangle

• They are the perpendicular bisectors of the sides of the outer triangle

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When we continue this discussion, we will get a few more interesting results:

Extend the three green lines in the fig.18.30(d) above. The extensions are shown by dashed green lines in fig.18.31(a) below:

Fig.18.31

We can see that the three lines meet at a point 'O'. Remember that these green lines are the perpendicular bisectors of the sides of the outer triangle ABC also. We have seen in the previous chapter on circles that, such perpendicular bisectors meet at a point, and the meeting point is the circumcentre.

So 'O' is the circumcentre of our outer triangle ABC. This is shown in the fig. 18.30(b). We can see that the circumcircle neatly passes through the three vertices A, B and C
We can write the following conclusion:
■ Two methods to obtain the circumcentre of any triangle ABC:
• Method 1: Draw the perpendicular bisector of any two sides. They will meet at the circumcentre
• Method 2: (i) Draw the inner triangle PQR by joining the midpoints of the sides (ii) Drop perpendiculars from any two vertices to the opposite sides. These perpendiculars will meet at the circumcentre.

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In the next section, we will see Medians.

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