1.
A ABC and A DBC are two isosceles triangles on
the same base BC and vertices A and D are on the
same side of BC (see Fig. 7.39). If AD is extended
to intersect BC at P, show that
(i) A ABD=AACD
(ii) A ABP=AACP
(iii) AP bisects Z A as well as D.
(iv) AP is the perpendicular bisector of BC.
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Appropriate Question:
Δ ABC and Δ DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see Fig. 7.39). If AD is extended to intersect BC at P, show that
(i) Δ ABD ≅Δ ACD
(ii) Δ ABP ≅ Δ ACP
(iii) AP bisects ∠ A as well as ∠ D.
(iv) AP is the perpendicular bisector of BC.
Given that, Δ ABC and Δ DBC are two isosceles triangles on the same base BC.
Now, Consider
Now, Consider
Now,
Thus, we concluded that
Now, Consider
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Please find the explanation in the attached photograph.
Step-by-step explanation:
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