ΔAMT ~ ΔAHE. In ΔAMT, AM = 6.3 cm, ∠TAM = 50°, AT = 5.6 cm. AM/AH = 7/5. Construct ΔAHE.

Answer: In ΔAHE, AH = 4.5 cm, ∠EAH = 50°, AE = 4 cm.

Analysis: \(\dfrac{AM}{AH}=\dfrac{AT}{AE}=\dfrac{MT}{HE}=\dfrac75\) and \(\angle TAM=\angle EAH=50^\circ\). ΔAHE is smaller; A is common, so H lies on seg AM and E on seg AT.

\(AH = 6.3\times\tfrac57 = 4.5\) cm, \(AE = 5.6\times\tfrac57 = 4\) cm.

Steps of construction:

1. Draw seg AM = 6.3 cm. At A draw \(\angle MAT = 50^\circ\) and mark T with AT = 5.6 cm. Join MT. This is ΔAMT.
2. Draw ray AX making an acute angle with AM, on the side opposite to T. Mark 7 equal parts \(A_1,\dots,A_7\).
3. Join \(A_7M\). Through \(A_5\) draw a line parallel to \(A_7M\), meeting seg AM at H.
4. Through H draw a line parallel to MT, meeting seg AT at E.
5. ΔAHE is the required triangle.