Question

$△AMT\sim △ANE\cdot \ln △$ $AMT,AM=6.3cm,\angle TAM=$ ${50}^{\circ },AT=5.6cm\cdot \frac{AM}{AH}=\frac{7}{5}$ Construct $△AHE.$

Answer 1

Analysis:
As shown in the figure,
Let $A-H-M$ and $A-E-T$.
$△AMT\sim △AHE.$ [Given]
$\therefore \angle TAM\cong \angle EAH$... [Corresponding angles of similar triangles]$\frac{AM}{AH}=\frac{MT}{HE}=\frac{AT}{AE}\dots .$ (i) [Corresponding sides of similar triangles]
But, $\frac{AM}{AH}=\frac{7}{5}\dots$ (ii)[Given]
$\therefore \frac{AM}{AH}=\frac{MT}{HE}=\frac{AT}{AH}=\frac{7}{5}\dots [$ From (i) and (ii)]
$\therefore$ Sides of AAMT are longer than corresponding sides of $△$ AHE.
$\therefore$ The length of side $AH$ will be equal to 5 parts out of 7 equal parts of side AM.

So, if we construct AAMT, point $H$ will be on sideAM, at a distance equal to 5 parts from $A$. Now, point $E$ is the point of intersection of ray AT and a line throughH, parallel to MT. $△$ $AHE$ is the required triangle similar to $△AMT$.
Steps of construction:
i. Draw $△AMT$ of given measure. Draw ray $AB$ making an acute angle with side AM.
ii. Taking convenient distance on the compass, mark 7 points $A1,A2,A3,A4,A5,Ag$ and $A7$, such that ${AA}_1=A_1{A}_2=$ $A_2{A}_3=A_3{A}_4=A_4{A}_5=A_5{A}_6=A_6{A}_7$
iii. Join $A_{7}M$. Draw line parallel to $A_{7}M$ through $A_5$ to intersects seg $AM$ at $H_{\text{s}}$
iv. Draw a line parallel to side TM through $H$. Name the point of intersection of this line and seg AT as $E_{\text{.}}$
$△AHE$ is the required triangle similar to $△AMT$.