Question
In following fig. ABC is an equilateral triangle . A circle is drawn with centre A so that ot cuts AB and AC at M and N respectively. Prove that BN = CM.
✓
Answer
ABC is an equilateral triangle,
∴ AB = AC
Also AN = MB (radii of same circle)
⇒ NC = MB
In Δ BNC and Δ CMB
NC = MB (proved above)
∠ B = ∠ C (60° each)
BC = BC (common)
∴ Δ BNC and Δ CMB (SAS)
∴ BN = CM (CPCT)
∴ AB = AC
Also AN = MB (radii of same circle)
⇒ NC = MB
In Δ BNC and Δ CMB
NC = MB (proved above)
∠ B = ∠ C (60° each)
BC = BC (common)
∴ Δ BNC and Δ CMB (SAS)
∴ BN = CM (CPCT)
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