Question

AB is a line segment and P is its mid-point. D and E are points on the same side of AB such that ∠ BAD = ∠ ABE and ∠ EPA = ∠ DPB (see Fig). Show that 

(i) ∆ DAP ≅ ∆ EBP 

(ii) AD = BE

Answer 1

It is given that: 

\[ \angle EPA = \angle DPB \]

Using the angle addition property:

\[ \angle EPA + \angle DPE = \angle DPB + \angle DPE \]

Therefore, we have:

\[ \angle DPA = \angle EPB \]

In \( \triangle DAP \) and \( \triangle EBP \):

\[ \angle DAP = \angle EBP \quad \text{(Given)} \]

\[ AP = BP \quad \text{(P is the mid-point of AB)} \]

\[ \angle DPA = \angle EPB \quad \text{(From above)} \]

By the ASA congruence rule:

\[ \triangle DAP \cong \triangle EBP \]

By CPCT (Corresponding Parts of Congruent Triangles):

\[ AD = BE \]