In △ ABC, ab = AC, the circle O with ab as diameter intersects BC at point D and AC at point e. it is proved that arc BD = arc De

In △ ABC, ab = AC, the circle O with ab as diameter intersects BC at point D and AC at point e. it is proved that arc BD = arc De

prove:
Connect ad
∵ AB is the diameter of ⊙ o
∴AD⊥BC
∵AB=AC
∴∠BAD=∠CAD
Ψ arc BD = Arc de

In a triangle ABC, BD ⊥ AC, ab ⊥ EC, compare the size of AB + EC and AC + BD, and explain the reason
Time doesn't wait. It's due tomorrow
AB＜AC

Lemma: ordering inequality, if 0ec, AB > BD (the hypotenuse is larger than the right angle), then Max {AB, AC, BD, EC} = max {AB, AC}, that is, if ab

In the triangle ABC, a = 3. B = root 7. C = 2?

∵cosB=（a²+c²-b²）/2ac=（9+4-7）/2*3*2=1/2
The angle is 60 ° or 120 °
In a triangle, the big angle is opposite to the big side. If angle B is 120 degrees, then angle a is larger, so it is 60 degrees

In the triangle ABC, a = root 7, B = 2, C = 3, then angle a =?

Using cosine theorem:
cos∠A=(b^2+c^2-a^2)/2bc=(4+9-7)/(2*2*3=6/12=1/2.
∴∠A=60°