Mathematics
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.
Answer
In △ ADC,
⇒ ∠ADB = ∠1 + ∠C [In a triangle an exterior angle is equal to the sum of two opposite interior angles.] ………(1)
In △ ADB,
⇒ ∠ADC = ∠2 + ∠B [In a triangle an exterior angle is equal to the sum of two opposite interior angles.] ………(2)
In △ ABC,
⇒ AC > AB (Given)
⇒ ∠B > ∠C [If two sides of a triangle are unequal, the greater side has the greater angle opposite to it.]
Since, AD is the bisector of angle A.
∴ ∠2 + ∠B > ∠1 + ∠C ………(3)
From equations (1), (2) and (3), we get :
⇒ ∠ADC > ∠ADB.
Hence, proved that ∠ADC is greater than ∠ADB.
Related Questions
In the following figure; AB is the largest side and BC is the smallest side of triangle ABC.
Write the angles x°, y° and z° in ascending order of their values.
In quadrilateral ABCD, side AB is the longest and side DC is the shortest. Prove that :
(i) ∠C > ∠A
(ii) ∠D > ∠B
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.
