Mathematics
Through any point in the bisector of an angle a straight line is drawn parallel to either arm of the angle. Prove that the triangle so formed is isosceles.
Triangles
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Answer
From figure,
AL is the bisector of angle A. Let D be any point on AL. From D, a straight line DE is drawn parallel to AC.
DE || AC
⇒ ∠ADE = ∠DAC (Alternate angles are equal) …….(1)
⇒ ∠DAC = ∠DAE (As, AL is the bisector of ∠A) ……(2)
From equation (1) and (2), we get :
⇒ ∠ADE = ∠DAE.
∴ AE = ED (Sides opposite to equal angles are equal)
∴ AED is an isosceles triangle.
Hence, proved that triangle formed is 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.
