AD is the median of the triangle ABC. If P is any point on AD, then which one of the following is correct?

Question

Prepp Team · Accepted Answer

Understanding Triangle Areas and Medians

The question asks about the relationship between the areas of triangle PAB and triangle PAC, given that AD is the median of triangle ABC and P is any point on AD.

Let's break this down using the property of a median in a triangle.

Property of a Median

A median of a triangle is a line segment joining a vertex to the midpoint of the opposite side. An important property of a median is that it divides the triangle into two triangles of equal area.

Applying the Median Property to Triangle ABC

Given that AD is the median of triangle ABC, D is the midpoint of BC. According to the property of a median, AD divides triangle ABC into two triangles of equal area:

Triangle ABD
Triangle ACD

Therefore, we can state:

\(\text{Area}(\triangle\text{ABD}) = \text{Area}(\triangle\text{ACD})\)

Let's call this Equation 1.

Applying the Median Property to Triangle PBC

Now consider triangle PBC. Since AD is the median of triangle ABC, D is the midpoint of BC. P is any point on AD. The line segment PD connects vertex P to the midpoint D of the opposite side BC in triangle PBC. This means PD is the median of triangle PBC.

Applying the same property of a median to triangle PBC, PD divides triangle PBC into two triangles of equal area:

Triangle PBD
Triangle PCD

Therefore, we can state:

\(\text{Area}(\triangle\text{PBD}) = \text{Area}(\triangle\text{PCD})\)

Let's call this Equation 2.

Relating Areas of Triangle PAB and Triangle PAC

Now let's look at how triangles PAB and PAC are formed.

Triangle PAB can be seen as the difference between triangle ABD and triangle PBD:

\(\text{Area}(\triangle\text{PAB}) = \text{Area}(\triangle\text{ABD}) - \text{Area}(\triangle\text{PBD})\)

Similarly, triangle PAC can be seen as the difference between triangle ACD and triangle PCD:

\(\text{Area}(\triangle\text{PAC}) = \text{Area}(\triangle\text{ACD}) - \text{Area}(\triangle\text{PCD})\)

Conclusion

From Equation 1, we know \(\text{Area}(\triangle\text{ABD}) = \text{Area}(\triangle\text{ACD})\).

From Equation 2, we know \(\text{Area}(\triangle\text{PBD}) = \text{Area}(\triangle\text{PCD})\).

Since the minuends (Area(ABD) and Area(ACD)) are equal, and the subtrahends (Area(PBD) and Area(PCD)) are also equal, the results of the subtraction must be equal.

Therefore:

\(\text{Area}(\triangle\text{PAB}) = \text{Area}(\triangle\text{PAC})\)

This shows that if AD is the median of triangle ABC and P is any point on AD, the area of triangle PAB is equal to the area of triangle PAC.

Triangle	Area Calculation
Triangle PAB	Area(\(\triangle\)ABD) - Area(\(\triangle\)PBD)
Triangle PAC	Area(\(\triangle\)ACD) - Area(\(\triangle\)PCD)

Since Area(\(\triangle\)ABD) = Area(\(\triangle\)ACD) and Area(\(\triangle\)PBD) = Area(\(\triangle\)PCD), it logically follows that Area(\(\triangle\)PAB) = Area(\(\triangle\)PAC).

Revision Table: Key Concepts

Concept	Definition/Property	Relevance to Problem
Median of a Triangle	Line segment from a vertex to the midpoint of the opposite side.	AD is the median of \(\triangle\)ABC; PD is the median of \(\triangle\)PBC.
Area Property of Median	A median divides a triangle into two triangles of equal area.	Used to state Area(\(\triangle\)ABD) = Area(\(\triangle\)ACD) and Area(\(\triangle\)PBD) = Area(\(\triangle\)PCD).
Area Subtraction	Area of smaller triangle = Area of larger triangle - Area of overlapping/subtracted part.	Used to express Area(\(\triangle\)PAB) and Area(\(\triangle\)PAC) in terms of other triangle areas.

Additional Information: Medians and Area Division

The property that a median divides a triangle into two equal areas is fundamental in triangle geometry. This property holds true regardless of the type of triangle (acute, obtuse, right-angled) or the position of the median.

If all three medians of a triangle are drawn, they intersect at a single point called the centroid. The centroid divides each median in a 2:1 ratio (vertex to centroid is twice the length of centroid to midpoint). Furthermore, the three medians divide the triangle into six smaller triangles, all of which have equal areas. This is a more advanced application of the median property but stems from the same fundamental idea that a median bisects the area.

In this specific problem, the point P being on the median AD is crucial. If P were outside the median, the relationship between Area(PAB) and Area(PAC) would generally be different.