**Example 13.** In △DEW\triangle DEW, AB∥EWAB \parallel EW. If AD=4 cmAD = 4\text{ cm}, DE=12 cmDE = 12\text{ cm} and DW=24 cmDW = 24\text{ cm}, find the length of DBDB. *Figure Description:* Triangle DEWDEW with vertex DD at the top, base EWEW at the bottom, and a line segment ABAB parallel to EWEW intersecting side DEDE at point AA and side DWDW at point BB. **Example 14.** In △PQR\triangle PQR, ST∥QRST \parallel QR, PSSQ=35\frac{PS}{SQ} = \frac{3}{5} and PR=28 cmPR = 28\text{ cm}, find the length of PTPT. *Figure Description:* Triangle PQRPQR with vertex PP at the top, base QRQR at the bottom, and a line segment STST parallel to QRQR intersecting side PQPQ at point SS and side PRPR at point TT.

Answer: For Example 13, DB=8 cmDB = 8\text{ cm}. For Example 14, PT=10.5 cmPT = 10.5\text{ cm}.

Step-by-step solution

Step 1: State Basic Proportionality Theorem for Example 13

In △DEW\triangle DEW, since segment AB∥EWAB \parallel EW, by the Basic Proportionality Theorem (Thales's theorem), the line dividing the sides creates proportional segments, giving DADE=DBDW\frac{DA}{DE} = \frac{DB}{DW}.

Step 2: Calculate DB in Example 13

Substitute the given measurements DA=4 cmDA = 4\text{ cm}, DE=12 cmDE = 12\text{ cm}, and DW=24 cmDW = 24\text{ cm} into the ratio. Solving for DBDB, we find DB=8 cmDB = 8\text{ cm}.

Step 3: State Proportionality relation for Example 14

In △PQR\triangle PQR, since ST∥QRST \parallel QR, we apply the Basic Proportionality Theorem to relate the segments: PSSQ=PTTR\frac{PS}{SQ} = \frac{PT}{TR}. Since TR=PR−PTTR = PR - PT, we can express the entire right-hand side in terms of PTPT and PRPR.

Step 4: Substitute values and solve for PT in Example 14

Substitute PSSQ=35\frac{PS}{SQ} = \frac{3}{5} and PR=28 cmPR = 28\text{ cm}. Expanding the cross-multiplication gives 84−3PT=5PT84 - 3PT = 5PT, which simplifies to 8PT=848PT = 84, giving PT=10.5 cmPT = 10.5\text{ cm}.

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