英文等腰三角形數學題6(20分)

"<" means angle "*'' means congruent
a) <A=<A (common<)
CD=BE (given)
<ADC=<AEB=90 (given)
so ABE*ACD (AAS)

b)AB=AC (corr.side,*triangle)
so ABC is an isos. triangle

c)AB=AC (proved in (b) )
<ABC=<ACB (base < of isos. triangle)
<ABE=<ACD (corr.< , isos. triangle)
<ABC - <ABE=<ACB - <ACD
so<FBC=<FCB
so BCF is an isos. triangle

(a) Given that ∠CDA=∠BEA, also ∠BAE=∠CAD (common angle)
also given that CD=BE, thus ΔABE congruent to ΔACD (AAS)

(b) by (a), we have AB=AC (corr. sides, congruent Δs)
thus ΔABC is isos. Δ

(c) by (b), we have ∠ABC=∠CBA (base ∠, isos.Δ)
and by (a), we have ∠ABE=∠ACD (corr. ∠, congruent Δs)
substracts the second equation from the first one,
we have ∠FBC=∠FCB, and thus FB=FC (opp. sides, equal ∠s)
and finally ΔBCF is isos. Δ.