A regular pentagon $ABCDE$ has diagonals $AC$ and $BD$, which intersect at $F$. Show that $\Delta ABF$ is isosceles. Show hint

See what angles you can work out.

If we can show that $\angle ABF = \angle AFB$ then $\Delta ABF$ must be isosceles.

We know that the angles in a pentagon must add up to $540^\circ$. So a regular pentagon has internal angles of $540^\circ \div 5 = 108^\circ$ ($\angle ABC$ for instance).

Now perhaps we should try to work out the other angles in $\Delta ABC$. Because $AB = BC$, the triangle is isosceles, and so $\angle BAC = \angle ACB = \frac{1}{2}(180^\circ - 108^\circ) = 36^\circ$. Similarly, $\angle CBD = 36^\circ$.

Knowing this, we can find $\angle ABF$: \begin{align*} \angle ABF &= \angle ABC - \angle CBD\\ &= 108^\circ - 36^\circ\\ &= 72^\circ \end{align*} and $\angle AFB$ from the sum of the angles in $\Delta ABF$: \begin{align*} \angle AFB &= 180^\circ - \angle BAF - \angle ABF\\ &= 180^\circ - 36^\circ - 72^\circ\\ &= 72^\circ \end{align*} Therefore $\angle ABF = \angle AFB$ so $\Delta ABF$ is isosceles.

Another way to find $\angle AFB$ would have been to show that $\angle BFC = 108^\circ$ from the sum of the angles in $\Delta BCF$, and then take the supplement to find $\angle AFB$.