∆AMT ~ ∆AHE. In ∆AMT, AM = 6.3 cm, ∠TAM = 50°, AT = 5.6 cm. `"AM"/"AH" = 7/5`. Construct ∆AHE.
∆AMT ~ ∆AHE. In ∆AMT, AM = 6.3 cm, ∠TAM = 50°, AT = 5.6 cm. `"AM"/"AH" = 7/5`. then construct △AMT and ΔAHE.
Solution
Analysis:

As shown in the figure,
Let A – H – M as well as points A – E – T be collinear.
∆AMT ~ ∆AHE,
∴ ∠TAM ≅ ∠EAH ...(Corresponding angles of similar triangles)
`"AM"/"AH" = "MT"/"HE" = "AT"/"AE"` ...(i)(Corresponding sides of similar triangles)
∴ `"AM"/"AH" = 7/5` ...(ii)(Given)
`"AM"/"AH" = "MT"/"HE" = "AT"/"AE" = 7/5` ...[From (i) and (ii)]
∴ sides of ∆AHE are smaller than sides of ∆AMT.
∴ If seg AH will be equal to 5 parts out of 7 equal parts of side AM.
So, if we construct ∆AMT, point H will be on side AM, at a distance equal to 5 parts from A.
Now, point E is the point of intersection of ray AT and a line through H, parallel to MT.
∆AHE is the required triangle similar to ∆AMT.
Steps of Construction:
 Draw ∆AMT such that AM = 6.3 cm, ∠TAM = 50°, AT = 5.6 cm.
 Draw ray AB making an acute angle with side AM.
 Taking convenient distance on the compass, mark 7 points A_{1}, A_{2}, A_{3}, A_{4}, A_{5}, A_{6} and A_{7}, such that AA_{1} = A_{1}A_{2} = A_{2}A_{3} = A_{3}A_{4} = A_{4}A_{5 }= A_{5}A_{6 }= A_{6}A_{7}.
 Join A_{7}M. Draw line parallel to A_{7}M through A_{5} to intersects seg AM at H.
 Draw a line parallel to side TM through H. Name the point of intersection of this line and seg AT as E.
∆AHE is the required triangle similar to ∆AMT.
Here, ∆AMT ~ ∆AHE.