Question

Suppose given a trapezoid ABCD with bases BC=5 and AD=8, and leg CD=3. The lines AB and CD meet at a point P. Find DP

Answer 1

Answer:

8 units

Step-by-step explanation:

Consider triangles BPC and APD. These two triangles are similar, because

• ∠DAP≅∠CBP (as corresponding angles);
• ∠BCP≅∠ADP (as corresponding angles);
• ∠APD≅∠BPC (by reflexive property of congruence).

Then, by AA theorem, $\triangle BCP\sim \triangle ADP.$

Similar triangles have proportional sides lengths:

$\dfrac{BP}{AP}=\dfrac{CP}{DP}=\dfrac{BC}{AD},\\ \\\dfrac{BP}{AP}=\dfrac{CP}{CP+3}=\dfrac{5}{8}.$

The last proportion gives you

$8CP=5(CP+3),\\ \\8CP=5CP+15,\\ \\3CP=15,\\ \\CP=5\ un.$

Then $DP=CP+CD=5+3=8\ un.$

Answer 2

Answer:

8

Step-by-step explanation:

ΔPBC ~ ΔPAD so ...

... BC/AD = CP/DP

... 5/8 = (DP-3)/DP

... 5·DP = 8(DP -3) . . . . multiply  by 8·DP

... 24 = 3DP . . . . . . . . . add 24-5·DP

... 8 = DP . . . . . . . . . . . . divide by 3