The value of AB id 0.5 unit.
The Angle-Angle-Angle (AAA) criterion for the similarity of triangles states that “If in two triangles, corresponding angles are equal, then their corresponding sides are in the same ratio (or proportion) and hence the two triangles are similar”.
Any pair of angles each of which is on the same side of one of two lines cut by a transversal and on the same side of the transversal are called corresponding angles and they are always equal.
According to the given question.
A, B and C lie on the straight line segment. and A, E, and D are also lie in the straight line segment.
Also, BE = 1.75m, CD = 7m and BC = 1.5m
Now, in ΔAEB and ΔADC
EB is parallel to BC
So,
∠AEB ≅ ∠ADC (corresponding angles)
∠ABE ≅ ∠ACD (corresponding angles)
∠EAB ≅ ∠DAC (common angle to the both triangle)
Hence, by AAA Similarity rule
ΔAEB is similar to ΔADC
Therefore,
[tex]\frac{AE}{AD} =\frac{BE}{CD} =\frac{AB}{AC}[/tex] (sides of the similar triangles are in proportion)
So,
[tex]\frac{BE}{CD} =\frac{AB}{AC} \\\implies \frac{AB}{AC} =\frac{1.75}{7} ....(i)[/tex]
Now, AC = AB + BC (as, A, B, and C lie on a straight line)
[tex]\implies \frac{AB}{AB+BC} =\frac{1.75}{7}[/tex] (from (i))
[tex]\implies \frac{AB}{AB+1.5} =\frac{1.75}{7}[/tex]
[tex]\implies 7AB = 1.75AB + 1.75(1.5)[/tex]
[tex]\implies 7AB -1.75AB = 2.625[/tex]
[tex]\implies 5.25AB = 2.625[/tex]
[tex]\implies AB = \frac{2.625}{5.25} = 0.5[/tex]
Hence, the value of AB id 0.5 unit.
Find out more information about AAA similarity and corresponding angles here:
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