Mathematics
AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that
(i) AD bisects BC
(ii) AD bisects ∠A
Triangles
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Answer
Given :
Δ ABC is an isosceles triangle and AB = AC.
AD is altitude
∴ ∠ADB = ∠ADC = 90°.
(i) In Δ BAD and Δ CAD,
⇒ ∠ADB = ∠ADC (Each equal to 90° as AD is altitude)
⇒ AB = AC (Given)
⇒ AD = AD (Common)
∴ Δ BAD ≅ Δ CAD (By R.H.S. Congruence rule)
We know that,
Corresponding parts of congruent triangles are equal.
∴ BD = CD (By C.P.C.T.)
Hence, proved that AD bisects BC.
(ii) Since, Δ BAD ≅ Δ CAD
∴ ∠BAD = ∠CAD (By C.P.C.T.)
Hence, proved that AD bisects ∠A.
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