Question

In th fig ABC and AMP are two right triangles, right angled at B and M respectively. Prove that ABC = AMP,

Answer 1

In two triangles , if corresponding angles are equal , then their corresponding sides are in the same ratio i.e they are proportional & hence the two triangles are similar..

If two angles of one triangle are respectively equal to two angles of another triangle then the two Triangles are similar because by the angle sum property of a triangle their third angle will also be equal and it is called AA similarity CRITERION.

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Solution:

Given:

∠ABC = 90°  & ∠AMP = 90°

(i) In ΔABC and ΔAMP, we have

∠A = ∠A (common angle)

∠ABC = ∠AMP = 90° (each 90°)

∴  ΔABC ~ ΔAMP (By AA similarity criterion)

(ii) As, ΔABC ~ ΔAMP (By AA similarity criterion)

If two triangles are similar then the corresponding sides are equal,

Hence, CA/PA = BC/MP..

In ΔABC and ΔAMP, we have

∠A = ∠A (common angle)

∠ABC = ∠AMP = 90° (each 90°)

∴  ΔABC ~ ΔAMP (By AA similarity criterion)

(ii) As, ΔABC ~ ΔAMP (By AA similarity criterion)

If two triangles are similar then the corresponding sides are equal,

Hence, CA/PA = BC/MP