Lines $A B$ and $C D$ are parallel and $P$ is any point between the two lines as shown in figure. Show that $∠ D P B = x + y$

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Solution

$L M ∥ A B$ and $C D$
$∠ A B P = ∠ B P M = x$ ..... $(1)$ (alternate interior angles)
$∠ D P M = ∠ P D C = y$ ..... $(2)$ (alternate interior angles)
$∠ D P B = ∠ B P M + ∠ D P M$
From $(1)$ and $(2)$ we have
$∠ D P B = x + y$

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Lines AB and CD are parallel and P is any point between the two lines as shown in figure. Show that ∠DPB=x+y

LM∥AB and CD∠ABP=∠BPM=x .....(1)(alternate interior angles)∠DPM=∠PDC=y .....(2)(alternate interior angles)∠DPB=∠BPM+∠DPMFrom (1) and (2) we have∠DPB=x+y

Lines $A B$ and $C D$ are parallel and $P$ is any point between the two lines as shown in figure. Show that $∠ D P B = x + y$

$L M ∥ A B$ and $C D$
$∠ A B P = ∠ B P M = x$ ..... $(1)$ (alternate interior angles)
$∠ D P M = ∠ P D C = y$ ..... $(2)$ (alternate interior angles)
$∠ D P B = ∠ B P M + ∠ D P M$
From $(1)$ and $(2)$ we have
$∠ D P B = x + y$