Mathematics
Prove that the straight line joining the vertex of an isosceles triangle to any point in the base is smaller than either of the equal sides of the triangle.
Answer
Let ABC be the isosceles triangle with AB = AC.
We know that,
The exterior angles of a triangle is always greater than each of the interior opposite angles.
∴ In △ ABD,
∠ADC > ∠B …….(1)
In △ ABC,
AB = AC (Given)
∴ ∠B = ∠C ……….(2)
From equations (1) and (2), we get :
∠ADC > ∠C
In △ ADC,
∠ADC > ∠C (Proved above)
∴ AC > AD …….(3)
Since, AB = AC, substituting in equation (3), we get :
⇒ AB > AD …….(4)
Since, AB > AD and AC > AD.
Hence, proved that the straight line joining the vertex of an isosceles triangle to any point in the base is smaller than either of the equal sides of the triangle.
Related Questions
In the following figure, ABC is an equilateral triangle and P is any point in AC; prove that :
(i) BP > PA
(ii) BP > PC
P is any point inside the triangle ABC. Prove that : ∠BPC > ∠BAC.
In the following diagram; AD = AB and AE bisects angle A. Prove that :
(i) BE = DE
(ii) ∠ABD > ∠C
In the figure, AB = AC and D is any point on BC, show that AB > AD.
