Question

In the given fig. AB//CD and P is any point as shown. Prove that <ABP+<BPD+<CDP=360° :

Answer 1

at first draw the parralel line through p
now,
AB||EP
<ABP+<BPE=180°
and, EP||CD
<EPD+<CPD=180°
NOW, <ABP+<BPE+<EPD+<CPD=180°+180°
<ABP+<BPD+<CPD=360°(PROVED)

Answer 2

Answer:

∠ABP+∠BPD+∠CPD = 360

Step-by-step explanation:

AB║EP

∠ABP + ∠BPE = 180

EP║CD

∠EPD + ∠ CPD = 180

∠ABP + ∠BPE + ∠EPD + ∠CPD = 180 + 180

∠ABP + ∠BPD + ∠CPD = 360