IT IS KNOWN THAT B C AND D IN ABC ARE THE MIDPOINTS OF THE BC EDGE

<> solution: Let the midpoint of BC be d, the central angle of the circle is equal to half of the circumferential angle, BOD=60°, and in the right-angled triangle BOD, there is OD=1

ob=12, so the trajectory equation for the midpoint d is: x2+y24, so d is chosen

Proof that since bc cuts the minor circle o to d, od is perpendicular to bc to d, and bc is the chord of the great circle o, so the midpoint of bc is d (bisecting the chord perpendicular to the diameter of the string).

Hello!! 1) Proof: Extend the ED to AB at point F.

ead=∠bad，de⊥ad

af=ae,df=de

and d is the midpoint of BC.

bfd≌△ced

ec=bfab=af+bf=ae+ec

2) AD=1 2BC, D is the midpoint of BC, then BAC=90°, ABC+ ACB=90°

and abc= bce

ace=90°

The ace is a right-angled triangle.

(2) When BAC 90°, AB 8, AD 5, find the length of the line segment CE.

D is known to be the midpoint of BC, AD=5

So, bc = 2ad = 10 (midline on hypotenuse).

Known bac=90°, ab=8

So, from the Pythagorean theorem: ac=6

Applying the previous congruence directly to find the internal error parallelism, the complementary of the same side internal angles of the parallel lines proves ace=90

Let ce=x be known from the conclusion of the first question ae+ce=ab=8

So, ae=8-x

So, in RT ace there is the Pythagorean theorem that AC 2 + CE 2 = AE 2 i.e.: 6 2 + x 2 = (8-x) 2

solution, x=7 4

i.e., ce=7 4

Landlord, you misplaced the topic and categorized it.

Here is the English translation.

Proof: ACB 90°, RT ADC, 1 2 90°, AD CF, in RT EDC, 3 2 90°, get: 1 3. ①

fb ac, acb 90°, fbc 90°, get: fbc is a right angle.

ac=bc，③

From the above three formulas, we get: RT ADC RT FBC.

cd fb, known cd db, obtains: db fb.

From AC=BC, ACB 90°, we get: 4 45°, AB is the CBF bisector.

So, AB bisects DF perpendicularly (the three lines in an isosceles triangle are combined).

1) Take the special points of E and F, both of which are the midpoints of Ab and Ac, which can prove that abc and def are similar, so def is an isosceles triangle.

2) ab=2, ae=1, so the quadrilateral aedf area is 1