As shown in the figure, it is known that ab ‖ CD, BD bisection ∠ ABC intersects AC in O, CE bisection ∠ DCG. If ∠ ace = 90 °, please judge the position relationship between BD and AC and explain the reason

As shown in the figure, it is known that ab ‖ CD, BD bisection ∠ ABC intersects AC in O, CE bisection ∠ DCG. If ∠ ace = 90 °, please judge the position relationship between BD and AC and explain the reason

BD ⊥ AC. the reasons are as follows: ∵ ab ∥ CD, ∵ ABC = ∧ DCG, ∵ BD bisection ∧ ABC intersects AC with O, CE bisection ∧ DCG, ∧ abd = 12 ∧ ABC, ∧ DCE = 12 ∧ BCG, ∧ abd = ∧ DCE; ∫ ab ∥ CD, ∧ abd = ∧ D, ∧ d = ∧ DCE, ∧ BD ∧ CE, and ∧ ace = 90 ° and ∧ BD ⊥ AC

In △ ABC, ab = AC, extend AB to d so that BD = AB, CE is the middle line of △ ABC, connect CD, and verify ∠ ace = ∠ D

It is proved that △ ace is similar to △ ADC, ∵ CE is the middle line of △ ABC and ab = AC ∵ AE / AC = 1 / 2, and ∵ BD = ab ∵ AC / ad = 1 / 2 ∵ AE / AC = AC / AD and ∵ EAC = ∵ CAD ∵ ace is similar to △ ADC ∵ ace = ∵ D

It is known that in △ ABC, D is the midpoint of AB, e is the midpoint of CB, F and G are the trisection points of AC respectively, connecting DF and eg, and extending the intersection to H
It is known that in △ ABC, D is the midpoint of AB, e is the midpoint of CB, F and G are the trisection points of AC respectively, connecting DF and eg, and extending the intersection to H
Proof: the quadrilateral abch is a parallelogram

Connecting De, we can get de / / AC and de = 1 / 2Ac, so de / / FG and de = 3 / 2fg, so DF: FH = 1:2 is the midpoint of AC, which is also the midpoint of FG. Connecting Di, we can get fi: AF = 1:2, so di / / AHI is the midpoint of AC, D is the midpoint of AB, so di / / BC is ah / / BC. Similarly, we can prove AB / / CH, so abch is flat if two opposite sides are parallel

As shown in the figure, △ ABC, ab = AC, D is on BC (D is not at the midpoint of BC), de ⊥ AB is on e, DF ⊥ AC is on F, BG ⊥ AC is on G, proving: de + DF = BG

It is proved that the area of △ ABC = the area of △ abd + the area of △ ACD, 12ab · de + 12ac · DF = 12ac · BG, ∵ AB = AC, ∵ de + DF = BG