A straight line parallel to BC of ABC intersects AB and AC at points P and Q respectively. If AP = QC, PB = 4 units & AQ = 9 units .Then the length of AP isa)2.5 Unitsb)3 unitsc)6 unitsd)6.5 unitsCorrect answer is option 'C'. Can you explain this answer?

Question

Answer

Let the line parallel to BC intersect BC at point X. Then, by the parallel line theorem, we have:

AP/PB = AQ/QC

Since AP = QC, we also have:

AP/PB = AQ/QC = 1

Therefore, AP = PB and AQ = QC. Let's call this common value x. Then, we have:

AB = AP + PB = x + 4

AC = AQ + QC = x + x = 2x

BC = 4 + x + 2x = 4 + 3x

Now, let's use the triangle inequality on triangle ABC:

AB + AC > BC

(x + 4) + (2x) > 4 + 3x

3x + 4 > 3x + 4

This is true for any value of x. Therefore, the triangle inequality holds and triangle ABC is a valid triangle.

Now, let's use the fact that the line parallel to BC intersects AB and AC to find the lengths of BP and CQ. Let's call the point of intersection with AB Y and the point of intersection with AC Z. Then, we have:

BY = PB = 4

CZ = QC = x

Since the line is parallel to BC, we have:

XY/ZC = AB/AC

XY/(x + x) = (x + 4)/2x

Simplifying, we get:

XY = (x + 4)/2

Now, let's use the fact that BPY and CQZ are similar triangles to find the lengths of BP and CQ. We have:

BP/BY = CQ/CZ

BP/4 = CQ/x

Substituting XY = (x + 4)/2, we get:

BP/4 = CQ/x = (x + 4)/(2x)

Solving for x, we get:

x = 8/3

Therefore, we have:

AP = QC = x = 8/3

PB = 4

AQ = 2x = 16/3

BC = 4 + 3x = 16/3

Finally, we can use the Pythagorean theorem to find the length of AB and AC:

AB^2 = AP^2 + PB^2 = (8/3)^2 + 4^2 = 256/9

AC^2 = AQ^2 + QC^2 = (16/3)^2 + (8/3)^2 = 256/9

Therefore, AB = AC = sqrt(256/9) = 16/3.

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