Mathematics
Show that the perpendiculars drawn from the extremities of the base of an isosceles triangle to the opposite sides are equal.
Answer
Let ABC be an isosceles triangle with AB = AC.
CD and BE are perpendiculars drawn from extremities of base BC.
In △ADC and △AEB,
⇒ AC = AB (Given)
⇒ ∠AEB = ∠ADC (Both equal to 90°)
⇒ ∠A = ∠A (Common angle)
∴ △ADC ≅ △AEB (By A.A.S axiom)
⇒ BE = CD (Corresponding parts of congruent triangles are equal)
Hence, the perpendiculars drawn from the extremities of the base of an isosceles triangle to the opposite sides are equal.
Related Questions
In the given figure, side BA of △ABC has been produced to D such that CD = CA and side CB has been produced to E. If ∠BAC = 106° and ∠ABE = 128°, find ∠BCD.
In the given figure, AB = BC and AC = CD. Show that: ∠BAD : ∠ADB = 3 : 1
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
In an isosceles triangle, prove that the altitude from the vertex bisects the base.