isosceles triangles and their perpendiculars proof

Question

Let an isosceles triangle ABC in the Euclidean plane be given, with AB being the base. DRAW the perpendiculars AD and BE from A and B onto BC and AC. Show that

Answer 1

You may refer to the following figure:

$∠ADB=∠AEB=90^\circ\;(AD⊥BC,BE⊥AC)$

$∠DBA=∠EAB\;(base\;∠s, \, isos.Δ)$

$∴∠BAD=∠ABE\;(∠ \; sum \; of \; Δ)$

Add one more perpendicular $CF$ onto $AB$. $AD$, $BE$ and $CF$ intersect at one point, the orthocenter. Call the point $G$.

$∠AGF=∠CGD\;(vert. \; opp. \;∠s)$

$∠GFA=∠GDC=90^\circ\;(CF⊥AB)$

$∴ ΔAGF \sim ΔCGD \; (AAA)$

For the same reason, $ ΔBGF \sim ΔCGE$.

$∴ ∠ACF=∠ABE$ and $∠BCF=∠BAD \; (equal \; corr. \; ∠s)$.

$∴ ∠ABE=∠BAD=\frac{1}{2}∠ACB$