Mathematics
In the given figure, AD = AC and B is a point on CD produced, show that AB > AD.
Triangles
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Answer
In Δ ADC,
⇒ ∠ADC = ∠ACD [As, AC = AD and angle opposite to equal sides equal.]
We know that,
In an isosceles triangle, the base angles are acute as only one obtuse angle can exist in a triangle.
∴ ∠ADC and ∠ACD are acute angles.
From figure,
BDC is a straight line.
⇒ ∠ADB + ∠ADC = 180° [Linear pairs]
⇒ ∠ADB = 180° − ∠ADC
Since, ∠ADC is an acute angle thus, ∠ADB is an obtuse angle.
∴ In Δ ADB, ∠ADB is the largest angle.
∴ ∠ADB > ∠ABD
∴ AB > AD (If two angles of a triangle are unequal, the greater angle has the greater side opposite to it.)
Hence, proved that AB > AD.
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