Mathematics
A triangle ABC has ∠B = ∠C. Prove that :
(i) the perpendiculars from the mid-point of BC to AB and AC are equal.
(ii) the perpendicular from B and C to the opposite sides are equal.
Triangles
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Answer
△ ABC is shown in the figure below:
(i) From figure,
In △ BDE and △ CDF,
⇒ BD = CD (As D is the mid-point of BC)
⇒ ∠B = ∠C (Given)
⇒ ∠DEB = ∠CFD (Both equal to 90°)
∴ △ BDE ≅ △ CDF (By A.A.S. axiom)
We know that,
Corresponding parts of congruent triangles are equal.
∴ DE = DF.
Hence, proved that the perpendiculars from the mid-point of BC to AB and AC are equal.
(ii) Let perpendiculars from B and C touch sides AC and AB at point H and G.
From figure,
In △ ABH and △ ACG,
⇒ ∠AHB = ∠AGC (Both equal to 90°)
⇒ ∠BAH = ∠CAG (Common angle)
⇒ ∠HBA = ∠GCA (Since, two angles of a triangle are equal so third angle is also equal)
∴ △ ABH ≅ △ ACG (By A.A.A. axiom)
We know that,
Corresponding parts of congruent triangles are equal.
∴ BH = GC.
Hence, proved that the perpendicular from B and C to the opposite sides are equal.
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