In the given figure P and Q are the mid points of AC and AB. Also, PG

I have the same doubt, we are taking everything symmetrical here.

Also please explain on which similarity rule you made those two triangle similar.

The triangles do not have to be assumed to be any type of special triangle.

The question is testing two concepts:

(1) Mid point theorem

(2) Basic Proportionality Theorem

Which are related and similar:

(1st) midpoint theorem states that if we draw a line segment from the midpoint of one side of a triangle across to the midpoint of the opposite side of the triangle ———> then that segment will be the “Mid-segment” of the triangle.

The Mid-Segment will be:

Parallel to the 3rd side

And

Mid segment length = (1/2) * (length of that 3rd side)

This is exactly what is going on with line PQ

PQ is parallel to BC
And
PQ = (1/2) (length of BC) (which is important for the end

The basic proportionality theorem says that the line drawn parallel (line PQ) to one side of a triangle and drawn from mid-point to mid-point will create two similar triangles where:

AC / AP = AB / AQ = 2 / 1

The heights of the similar triangles will be in a 2/1 ratio as well

The same thing is occurring in triangle RPQ with line GH and line PQ

If we use PQ as the base of triangle PQR and BC as the base of triangle ABC we know that the ratio of bases will be:

(Base BC) / (Base PQ) = (2) / (1)

What chetan2u is doing is drawing a perpendicular height from vertex A to base BC and another perpendicular height from vertex R to base PQ.

Then he proves that these 2 heights are equal - since there exists only one perpendicular distance between 2 parallel lines, the parallel lines of PQ and BC

Since the heights drawn from their respective vertices are equal, the Area of the triangle ABC - to - Area of triangle PQR will be in the same Ratio as ———> (Base BC) / (Base PQ) = (2) / (1)

flipping it around to match the question the answer is

1/2

Note: this is a really hard question. I believe there are 2 official questions that test the Concept of the Midpoint Theorem (one DS and one PS)

I can’t imagine this question would be anything less than an upper 700 type level of question.




Posted from my mobile device

Since G and H are mid points of PR and QR, GH is parallel to PQ. This makes GRH and PRQ one of our standard similar triangle pairs such that sides are in the ratio 1:2. So if GH = a, PQ = 2a.
Also, altitudes will be in the ratio 1:2 too. So if altitude from R to GH is b, its rest of the length from GH to PQ is b too.

Similarly, APQ and ACB are our similar triangle pairs such that sides are in the ratio 1:2. So since PQ = 2a, BC = 4a.
Again, altitudes will be in the ratio 1:2 too. So if altitude from A to PQ is c, its rest of the length from PQ to GH will be c too.

So we see that the perpendicular length from GH to PQ is b and is also c. Hence b = c.
So the altitudes of PQR and ABC are equal. But their bases (PQ:BC) are in the ratio 1:2. Hence, their areas will be in the ratio 1:2 or we can say that area of PQR is 1/2 the area of ABC.

Answer (A)

For more, check out David's post here:
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2014/03 ... -trianges/
and also https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2014/0 ... -the-gmat/

Great question! Combines several concepts and theorems without requiring any involved calculation. Can be solved by reasoning only, however we must know the following concepts for a quick turnaround:

1. Mid-point theorem
2. Basic proportionality theorem
3. Similarity of triangles

The mid-point theorem is discussed in this video and the solution is discussed @ 3:06



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