In the figure, AB = AC and D is any point on BC, show that AB AD.

Question

KnowledgeBoat · Accepted Answer

In △ ABD and △ ACD, ⇒ AB = AC (Given) ⇒ ∠ABC = ∠ACB = θ (let) [Angles opposite to equal sides are equal] ⇒ AD = AD (Common side) ∴ △ ABD ≅ △ ACD (By S.A.S. axiom) We know that, Corresponding sides of congruent triangle are equal. ∴ ∠ADB = ∠ADC = x (let) From figure, ⇒ ∠ADB + ∠ADC = 180° ⇒ x + x = 180° ⇒ 2x = 180° ⇒ x = = 90°. ∴ ∠ADB = ∠ADC = 90°. We know that, The base angles of isosceles triangle are acute angles. ∴ θ 90° In △ ABD, ⇒ ∠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.

Mathematics

In the figure, AB = AC and D is any point on BC, show that AB > AD.

Triangles

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