$A B$ is a line segment and $P$ is its mid-point. $D$ and $E$ are points on the same side of $A B$ such that $∠ B A D = ∠ A B E$ and $∠ E P A = ∠ D P B$ . Show that $A D = B E$ .

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Solution

$∠ B A D = ∠ A B E \Rightarrow ∠ P A D = ∠ P B E$

$∠ E P A = ∠ D P B$

$\Rightarrow ∠ D P A + ∠ E P D = ∠ E P D + ∠ E P B$

$\Rightarrow ∠ D P A = ∠ E P B$

$∴$ In $Δ D A P$ and $Δ E B P$

$∠ P A D = ∠ P B E - (i)$

$A P = B P - (i i)$

$∠ D P A = ∠ E P B - (i i i)$

$∴ Δ D A P ≅ Δ E B P$ by ASA test of congruence
$∵$ Two triangle are congruent
$∴$ Corresponding sides should be equal
$A D = B E$ (proved)

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Question 7 (ii)
AB is a line segment and P is its mid-point. D and E are points on the same side of
AB such that $∠ B A D = ∠ A B E and ∠ E P A = ∠ D P B$ (See the given figure). Show that
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