In ABC,

In ABC,

knowable

If Mn is a straight line passing through point a, BD ⊥ Mn is in D, CE ⊥ Mn is in E, then BD = AE 1. If Mn is rotated around point a
Let Mn and BC intersect at point O, other conditions remain unchanged, are BD and AE still equal? Why? 2. Under the condition of 1., what is the relationship between CE, BD and de?

(1) It is proved that BD ⊥ Mn and D, EC ⊥ Mn and E,
Then △ abd and △ CEA are right triangles,
And ab = AC, so △ abd ≌ △ CEA,
BD = AE;
If Mn is rotated around point a and intersects with BC at point O,
Then BD and CE are perpendicular to Mn,
The two triangles are congruent,
So BD and AE sides are still equal;
(3) Because △ abd ≌ △ CEA, then BD = AE, ad = EC, so BD + EC = De

As shown in the figure, in △ ABC, ∠ BAC = 90 ° AB = AC, the straight line Mn passes through point a, through point B as BD ⊥ Mn in D, through C as CE ⊥ Mn in E. (1) verify: △ abd ≌ △ CAE; (2) if BD = 12cm, de = 20cm, calculate the length of CE

(1) It is proved that: ∵ BAC = 90 °, and ∵ BD ⊥ Mn, CE ⊥ Mn, ∵ CAD ⊥ ace = 90 °, and ∵ BDA = - AEC = 90 °, and ≌ bad = - ace, and ab = AC. in △ abd and △ CAE, ≌ BDA = - AEC ⊥ bad = - aceab = AC, ≌ abd ≌ △ CAE (AAS); (?)

It is known that in the triangle ABC, ab = AC, angle ABC = 90 degrees, an is any straight line passing through point a, BD is perpendicular to point D, CE is perpendicular to point E, can you explain BD + CE

"Can you explain BD + CE" should be "can you explain BD + CE = de"?
Because CE ⊥ an, BD ⊥ an
So △ abd and △ ace are right triangles
So ∠ bad ＋ abd = 90 °,
Because ∠ BAC = 90 degree
Therefore, CAE + bad = 90 degree
Therefore, abd = CAE
And because ∠ ADB = ∠ BAC = 90 ° AB = AC
So △ abd ≌ △ CAE (AAS)
So CE = ad, AE = BD
So BD + CE = AE + ad
So BD + CE = De
For reference! Jswyc