Mathematics
BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle ABC is isosceles.
Answer
Triangle ABC with BE and CF as equal altitudes is shown in the figure below:
Given :
BE is a altitude.
∴ ∠AEB = ∠CEB = 90°
CF is a altitude.
∴ ∠AFC = ∠BFC = 90°
Also, BE = CF.
In Δ BEC and Δ CFB,
⇒ ∠BEC = ∠CFB (Each equal to 90°)
⇒ BC = CB (Common)
⇒ BE = CF (Given)
⇒ Δ BEC ≅ Δ CFB (By R.H.S. congruence rule)
We know that,
Corresponding parts of congruent triangle are equal.
⇒ ∠BCE = ∠CBF (By C.P.C.T.)
As,
Sides opposite to equal angles of a triangle are equal.
∴ AB = AC.
Hence, proved that Δ ABC is an isosceles triangle.
Related Questions
Δ ABC and Δ DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC. If AD is extended to intersect BC at P, show that
(i) Δ ABD ≅ Δ ACD
(ii) Δ ABP ≅ Δ ACP
(iii) AP bisects ∠A as well as ∠D
(iv) 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 Δ PQR. 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.