Mathematics
In a triangle ABC, AB = AC, D and E are points on the sides AB and AC respectively such that BD = CE. Show that:
(i) △DBC ≅ △ECB
(ii) ∠DCB = ∠EBC
(iii) OB = OC, where O is the point of intersection of BE and CD.
Triangles
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Answer
(i) From figure,
BD = CE (Given)
BC = BC (Common)
∠DBC = ∠ECB (as AB = AC).
∴ △DBC ≅ △ECB by SAS axiom.
Hence, proved △DBC ≅ △ECB.
(ii) We know that corresponding parts of congruent triangle are equal.
∴ ∠DCB = ∠EBC.
Hence, proved that ∠DCB = ∠EBC.
(iii) As △DBC ≅ △ECB,
∠BDO = ∠CEO (By c.p.c.t.)
∠DOB = ∠EOC (Vertically opposite angles)
BD = CE (Given)
Hence, △BOD ≅ △EOC by ASA axiom.
We know that corresponding parts of congruent triangles are equal.
∴ OB = OC.
Hence, proved that OB = OC.
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Related Questions
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