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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

 (i) \small \Delta DAP\cong \Delta EBP

            

Answers (1)

best_answer

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

or                                                            \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

 

Posted by

Devendra Khairwa

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