Mathematics
In Δ ABC, AD is the perpendicular bisector of BC. Show that Δ ABC is an isosceles triangle in which AB = AC.
Answer
Given :
AD is the perpendicular bisector of BC.
∴ ∠ADB = ∠ADC = 90° and BD = DC.
In Δ ADC and Δ ADB,
⇒ AD = AD (Common side)
⇒ ∠ADC = ∠ADB (Each equal to 90°)
⇒ CD = BD (AD is the perpendicular bisector of BC)
∴ Δ ADC ≅ Δ ADB (By S.A.S. congruence rule)
We know that,
Corresponding parts of congruent triangles are equal.
∴ AB = AC (By C.P.C.T.)
Hence, proved that ABC is an isosceles triangle in which AB = AC.
Related Questions
In an isosceles triangle ABC, with AB = AC, the bisectors of ∠B and ∠C intersect each other at O. Join A to O. Show that :
(i) OB = OC
(ii) AO bisects ∠A
ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively see Fig. Show that these altitudes are equal.
ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal. Show that
(i) Δ ABE ≅ Δ ACF
(ii) AB = AC, i.e., ABC is an isosceles triangle
