Question 14 of 14advanced🔧 ApplyLong Answer4 marks

If AD and PM are medians of triangles ABC and PQR, respectively where ΔABC∼ΔPQR\Delta ABC \sim \Delta PQR, prove that AB/PQ=AD/PMAB/PQ = AD/PM.

Correct Answer

Given: ΔABC∼ΔPQR\Delta ABC \sim \Delta PQR. AD and PM are medians of ΔABC\Delta ABC and ΔPQR\Delta PQR respectively.

To Prove: AB/PQ=AD/PMAB/PQ = AD/PM

Proof:

Since ΔABC∼ΔPQR\Delta ABC \sim \Delta PQR (Given),

ABPQ=BCQR=CARP... (i)\frac{AB}{PQ} = \frac{BC}{QR} = \frac{CA}{RP} \quad \text{... (i)}

and ∠B=∠Q\angle B = \angle Q, ∠C=∠R\angle C = \angle R ... (ii)

Since AD is median of ΔABC\Delta ABC, D is midpoint of BC.

So, BD=DC=BC2BD = DC = \frac{BC}{2}

Since PM is median of ΔPQR\Delta PQR, M is midpoint of QR.

So, QM=MR=QR2QM = MR = \frac{QR}{2}

From (i), BCQR=ABPQ\frac{BC}{QR} = \frac{AB}{PQ}

Therefore,

BDQM=BC/2QR/2=BCQR=ABPQ... (iii)\frac{BD}{QM} = \frac{BC/2}{QR/2} = \frac{BC}{QR} = \frac{AB}{PQ} \quad \text{... (iii)}

Now, in ΔABD\Delta ABD and ΔPQM\Delta PQM:

BDQM=ABPQ[from (iii)]\frac{BD}{QM} = \frac{AB}{PQ} \quad \text{[from (iii)]} ∠B=∠Q[from (ii)]\angle B = \angle Q \quad \text{[from (ii)]}

Therefore, by SAS similarity criterion, ΔABD∼ΔPQM\Delta ABD \sim \Delta PQM

Hence,

ABPQ=ADPM\frac{AB}{PQ} = \frac{AD}{PM}

(Corresponding sides of similar triangles are proportional)

Hence proved.

Exercise: EXERCISE 6.3 | Q: 16 | (Page Page 25)
For More Understanding

Explanation

This question tests understanding of properties of similar triangles and medians. The key insight is that when two triangles are similar, their corresponding medians are also proportional to corresponding sides.

The proof uses the SAS similarity criterion by showing that ΔABD\Delta ABD and ΔPQM\Delta PQM have proportional sides (AB/PQ=BD/QMAB/PQ = BD/QM) and equal included angles (∠B=∠Q\angle B = \angle Q). The textbook context confirms that corresponding angles are equal when triangles are similar, and medians divide the opposite side into two equal parts.

Solution Steps

  1. Step 1: Use the property of similar triangles to establish AB/PQ=BC/QRAB/PQ = BC/QR and ∠B=∠Q\angle B = \angle Q

  2. Step 2: Express BD and QM in terms of BC and QR using median property

  3. Step 3: Show that BD/QM=AB/PQBD/QM = AB/PQ

  4. Step 4: Apply SAS similarity criterion to prove ΔABD∼ΔPQM\Delta ABD \sim \Delta PQM

  5. Step 5: Conclude AB/PQ=AD/PMAB/PQ = AD/PM using corresponding sides property

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