Mathematics
Answer
Let ABC be an isosceles triangle with AB = AC.
Let AD be a perpendicular from vertex A to base BC.
In △ADB and △ADC,
⇒ AD = AD (Common side)
⇒ AB = AC (Given)
⇒ ∠ADB = ∠ADC (Each equal to 90°)
∴ △ADB ≅ △ADC (By R.H.S axiom)
∴ DB = DC (Corresponding parts of congruent triangles are equal)
Hence, proved that the altitude from the vertex in an isosceles triangle bisects the base.
Related Questions
Show that the perpendiculars drawn from the extremities of the base of an isosceles triangle to the opposite sides are equal.
In a △ABC, AB = AC. If the bisectors of ∠B and ∠C meet AC and AB at points D and E respectively, show that :
(i) △DBC ≅ △ECB
(ii) BD = CE
If the altitude from one vertex of a triangle bisects the opposite side, prove that the triangle is isosceles.
In the given figure, AD = AE and ∠BAD = ∠CAE. Prove that : AB = AC.
