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  • ∆ ABC and ∆ DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see Fig). If AD is extended to intersect BC at P, show that (ii) ∆ ABP≅ ∆ ACP

1.∆ ABC and ∆ DBC are two isosceles triangles on the same base BC and vertices A  and D are on the same side of BC (see Fig). If AD is extended to intersect BC at P, show that \small \Delta ABP \cong \Delta ACP

   

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Consider \Delta ABP  and   \Delta ACP,

(i)  AP  is common side in both the triangles.

(ii)  \angle PAB\ =\ \angle PAC (This is obtained from the c.p.c.t. as proved in the previous part.)

(iii)   AB\ =\ AC        (Isosceles triangles)

Thus by SAS axiom, we can conclude that :

    \small \Delta ABP \cong \Delta ACP

Posted by

Devendra Khairwa

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