The △ ACB is an isosceles triangle, ∠ ACB = 90 ° extend Ba to e, extend AB to F, ∠ ECF = 135 ° Q: is there an equivalent relationship between AB, AE and BF

The △ ACB is an isosceles triangle, ∠ ACB = 90 ° extend Ba to e, extend AB to F, ∠ ECF = 135 ° Q: is there an equivalent relationship between AB, AE and BF

It is proved that: because ∠ ECF = 135, ∠ ACB = 90, so ∠ ECF - ∠ ACB = 45 ° that is ∠ ECA + ∠ BCF = 45 ° because ∠ ACB = 45 ° that is ∠ BCF + ∠ f = 45 ° that is ∠ ECA = ∠ F, similarly ∠ e = ∠ BCF so △ ace ∽ BFC so AC / BF = AE / BC that is AC × BC = AE × BF, because in isosceles right triangle ACM, AC = √ 2am, in isosceles right triangle ABC, BC = (√ 2 / 2) AB so AC × BC = am × AB that is am × ab = AE × BF