Answer:
Step-by-step explanation:
To find the angles of ∠MBE and ∠SAE, given AE is the tangent line of circle S and circle M, we have to understand this rule:
"The tangent line of a circle is perpendicular to the radius that passes through the tangent point."
Therefore:
Now, we compare ΔMBE and ΔSAE. Both triangles shares 2 pairs of same angles, which are:
Hence, we can conclude that their 3rd angles must be the same ⇒ ∠BME = ∠ASE. Since both ΔMBE and ΔSAE have same corresponding angles, then ΔMBE and ΔSAE are similar triangles.
For similar triangles, the corresponding sides are in the same ratio:
[tex]\boxed{BE : AE = MB : SA = ME : SE}[/tex]
If SE = 100, ME = 5 and MB = 2, then:
[tex]\begin{aligned}MB:SA&=ME:SE\\2:SA&=5:100\\5\times SA&=2\times100\\SA&=200\div5\\\bf SA&=\bf40\end{aligned}[/tex]