# BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle ABC is isosceles.

**Solution:**

Let's construct a diagram according to the given question as shown below.

In ΔBEC and ΔCFB,

∠BEC = ∠CFB (Each 90°)

BC = CB (Common)

BE = CF (altitudes are equal given)

∴ ΔBEC ≅ ΔCFB (By RHS congruency)

∴ ∠BCE = ∠CBF (By CPCT)

∴ AB = AC (Sides opposite to equal angles of a triangle are equal)

Hence, ΔABC is isosceles.

**☛ Check: **NCERT Solutions for Class 9 Maths Chapter 7

**Video Solution:**

## BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle ABC is isosceles

NCERT Maths Solutions Class 9 Chapter 7 Exercise 7.3 Question 4:

**Summary:**

If BE and CF are two equal altitudes of a triangle ABC, then using the RHS congruence rule, we can prove that the triangle ABC is isosceles.

**☛ 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.
- AD is an altitude of an isosceles triangle ABC in which AB = AC.Show that i) AD bisects BC ii) AD bisects ∠A.
- 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
- ABC is an isosceles triangle with AB = AC. Draw AP ⊥ BC to show that ∠B = ∠C.

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