CLASS 10 MATHS CHAPTER 3 EXERCISE 3.3 KE QUESTION 1 KA 6 .?

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Answer

Explanation of Class 10 Maths Chapter 3 Exercise 3.3 Question 1 (6)

Given: A line segment AB with midpoint O is parallel to the line segment CD with midpoint P. Q is the midpoint of the line segment AC. PQ intersects AB at point M.

To prove: AQ = 2PM

Proof:

As line AB is parallel to line CD, angle ACD and angle BDC are alternate angles and hence are equal.

Similarly, angle ABC and angle ADC are alternate angles and hence are equal.

Thus, we have two pairs of equal angles which implies that triangles ABC and ADC are similar by AA similarity criterion.

Therefore, we have AB/AD = AC/AB

Given that Q is the midpoint of AC, we have AQ = QC

Also, O is the midpoint of AB, we have PM = MO

From the similarity of triangles ABC and ADC, we have AB/AD = AC/AB, which can be written as AB^2 = AC × AD.

Now, using the midpoint theorem in triangle ADC, we have PM = 1/2DC.

Also, using the midpoint theorem in triangle ABC, we have AQ = 1/2AC.

Substituting AC = 2AQ and DC = 2PM in AB^2 = AC × AD, we get AB^2 = 4AQ × AD/2.

Therefore, AB^2 = 2AQ × AD.

Now, using the Pythagoras theorem in triangle ABD, we have AB^2 = AD^2 + BD^2.

Substituting AB^2 = 2AQ × AD, we get 2AQ × AD = AD^2 + BD^2.

So, BD^2 = AD^2 + 2AQ × AD = AD(AD + 2AQ).

Since Q is the midpoint of AC, we have AD = 2AQ.

Thus, BD^2 = AD(AD + 2AQ) = 2AQ(2AQ + 2AQ) = 8AQ^2.

Therefore, BD = 2AQ.

Since M is the midpoint of PQ, we have PM = MQ.

Thus, PM = MQ = 1/2PQ.

Also, PQ = BD by the midpoint theorem in triangle ABC.

Therefore, PM = MQ = 1/2BD = 1/2(2AQ) = AQ.

Hence, AQ = 2PM, which was to be proved.

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