Mathematics
In a △ ABC, the internal bisector of angle A meets opposite side BC at point D. Through vertex C, line CE is drawn parallel to DA which meets BA produced at point E. Show that △ ACE is isosceles.
Triangles
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Answer
△ ABC is shown in the figure below:
Given,
DA || CE
⇒ ∠1 = ∠4 (Corresponding angles are equal) ……….(1)
⇒ ∠2 = ∠3 (Alternate angles are equal) …………..(2)
⇒ ∠1 = ∠2 (Since, AD is bisector of angle A) ………..(3)
Substituting value of ∠1 and ∠2 from equations (1) and (2) in equation (3), we get :
⇒ ∠4 = ∠3
⇒ ∠AEC = ∠ACE
In △ AEC,
⇒ AC = AE (Sides opposite to equal angles are equal)
Hence, proved that ACE is an isosceles triangle.
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Related Questions
In quadrilateral ABCD, side AB is the longest and side DC is the shortest. Prove that :
(i) ∠C > ∠A
(ii) ∠D > ∠B
In triangle ABC, side AC is greater than side AB. If the internal bisector of angle A meets the opposite side at point D, prove that : ∠ADC is greater than ∠ADB.
In isosceles triangle ABC, sides AB and AC are equal. If point D lies in base BC and point E lies on BC produced (BC being produced through vertex C), prove that :
(i) AC > AD
(ii) AE > AC
(iii) AE > AD
Given : ED = EC
Prove : AB + AD > BC.
