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Show that AE=AF for a=40^ exto.

April 30, 2026Mathematics

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AE=AF for a=40^ exto

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1b) Show that $AE=AF$ for $a=40^\circ$ .

Calculate the measure of $\angle AEF$ using the given value $a=40^\circ$ .
$\angle AEF = 2a = 2(40^\circ) = 80^\circ$
Compare $\angle AEF$ with $\angle AFE$ .
From the diagram, $\angle AFE = 80^\circ$ .
We found $\angle AEF = 80^\circ$ .
Conclude based on the angles.
Since $\angle AEF = \angle AFE = 80^\circ$ , $\triangle AEF$ is an isosceles triangle. In an isosceles triangle, the sides opposite the equal angles are equal.
Therefore, $\boxed{AE=AF}$ .

Regarding questions 2 and 3, they are not visible in the provided image. Please send the full questions if you would like me to solve them.

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