The following figure shows a triangle ABC in which AB = AC. M is a point on AB and N is a point on AC such that BM = CN. Prove that :

(i) AM = AN

(ii) △ AMC ≅ △ ANB

(iii) BN = CM

(iv) △ BMC ≅ △ CNB

(i) Given,

AB = AC = x (let) and BM = CN = y (let)

From figure,

⇒ AM = AB - BM = x - y

⇒ AN = AC - CN = x - y

∴ AM = AN.

Hence, proved that AM = AN.

(ii) In △ AMC and △ ANB,

⇒ AM = AN (Proved above)

⇒ ∠MAC = ∠NAB (Common angle)

⇒ AC = AB (Given)

∴ △ AMC ≅ △ ANB (By S.A.S. axiom).

Hence, proved that △ AMC ≅ △ ANB.

(iii) We know that,

Corresponding parts of congruent triangles are equal.

Since,

△ AMC ≅ △ ANB

∴ CM = BN.

Hence, proved that BN = CM.

(iv) We know that,

Angles opposite to equal sides are equal.

Since,

AB = AC

∴ ∠C = ∠B.

In △ BMC and △ CNB,

⇒ BM = CN (Given)

⇒ BC = BC (Common side)

⇒ ∠B = ∠C (Proved above)

∴ △ BMC ≅ △ CNB (By S.A.S. axiom).

Hence, proved that △ BMC ≅ △ CNB.