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  • 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

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

     

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From the figure, it is clear that : \angle EPA\ =\ \angle DPB

Adding \angle DPE both sides, we get : \angle EPA\ +\ \angle DPE =\ \angle DPB\ +\ \angle DPE                                                           

\angle DPA =\ \angle EPB

Now, consider \Delta DAP and   \Delta EBP :

(i) \angle DPA =\ \angle EPB                  (Proved above)

(ii) AP\ =\ BP                        (Since P is the midpoint of line AB)    

(iii) \small \angle BAD=\angle ABE                  (Given)

Hence by ASA congruence, we can say that :  \small \Delta DAP\cong \Delta EBP

 

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Devendra Khairwa

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