AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that
(i) AD bisects BC (ii) AD bisects ∠A
Given: AB = AC
Let's construct an isosceles triangle ABC in which AB = AC as shown below.
(i) In ΔBAD and ΔCAD,
∠ADB = ∠ADC (Each 90° as AD is an altitude)
AB = AC (Given)
AD = AD (Common)
∴ ΔBAD ≅ ΔCAD (By RHS Congruence rule)
∴ BD = CD (By CPCT)
Hence, AD bisects BC.
(ii) Since, ΔBAD ≅ ΔCAD
By CPCT, ∠BAD = ∠CAD
Hence, AD bisects ∠A.
AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that (i) AD bisects BC (ii) AD bisects ∠A
NCERT Maths Solutions Class 9 Chapter 7 Exercise 7.3 Question 2
If AD is an altitude of an isosceles triangle ABC in which AB = AC, then AD bisects BC and AD bisects ∠A.
☛ Related Questions:
- ΔABC and ΔDBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see the given figure). If AD is extended to intersect BC at P, show thati) ΔABD ≅ ΔACDii) ΔABP ≅ ΔACPiii) AP bisects ∠A as well as ∠Div) AP is the perpendicular bisector of BC.
- Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of ΔPQR (see Fig. 7.40). Show that:(i) Δ ABM ≅ Δ PQN(ii) Δ ABC ≅ Δ PQR
- BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle ABC is isosceles.
- ABC is an isosceles triangle with AB = AC. Draw AP ⊥ BC to show that ∠B = ∠C.