In the triangle ABC, ab = 15, AC = 20, the height ad on the side of BC = 12, try to find the length of BC? I give bonus points to the detailed process, get 7 within the 25 do not talk about I want the process of 7

In the triangle ABC, ab = 15, AC = 20, the height ad on the side of BC = 12, try to find the length of BC? I give bonus points to the detailed process, get 7 within the 25 do not talk about I want the process of 7

Analysis: two cases,
When high ad is on the extension line of CB,
In RT △ abd, AB ^ 2 = ad ^ 2 + BD ^ 2,
So BD ^ 2 = 15 ^ 2-12 ^ 2 = 81,
∴BD=9,
In RT △ ACD, AC ^ 2 = ad ^ 2 + CD ^ 2,
We get CD ^ 2 = 20 ^ 2-12 ^ 2 = 16 ^ 2,
∴CD=16,
Then BC = cd-bd = 16-9 = 7,
In essence, this case is an obtuse triangle. Another case is an acute triangle

In the triangle ABC, D is the midpoint of BC, ab = 5, ad = 6, AC = 13. Try to judge the position relationship between AB and AD
Let's extend ad to e, make de = ad, connect CE, and prove that ACE is a right triangle, AC ^ 2 = AE ^ 2 + CE ^ 2, and AEC is a right angle. If the triangle Dec and DAB are congruent, then the angle DAB is a right angle,
Ad and ab are perpendicular.

Vertical
Let BD = DC = X
From cosine theorem
There is CoSb = (AB ^ 2 + BD ^ 2-ad ^ 2) / 2Ab * BD in triangle abd
There is CoSb = (AB ^ 2 + BC ^ 2-ac ^ 2) / 2Ab * BC in triangle ABC
Substituting AB = 5, AC = 13, ad = 6, BD = x, BC = 2x into the solution, we get x = radical 61
So AB ^ 2 + ad ^ 2 = 5 ^ 2 + 6 ^ 2 = x ^ 2
So the triangle abd is a right triangle, and the angle bad = 90 degrees, AB and AD are perpendicular

In the triangle ABC, AB is equal to 5, and the middle line ad on the side of BC is equal to 4?

In the triangle ABC, AB is equal to 5, and the middle line ad on the side of BC is equal to 4. Find the length of AC
It also needs the length of the street
AD²+（BC/2）²-AB²+AD²+（BC/2）²-AC²=0

In the triangle ABC, AB equals 13, AC equals 20, high ad equals 12, then the length of BC is () process!
2. The square sum of three sides is 1800 square centimeter and the hypotenuse is ()

1、
BC=√(13²-12²)+√(20²-12²)
=5+16
=21
2、
a²+b²=c²
a²+b²+c²=1800
So 2C & # 178; = 1800
c²=900
C = 30 cm

As shown in the figure, in △ ABC, ab = AC = 20, BC = 32, point D is on BC, and ad = 13

As shown in the figure, AE ⊥ BC is e through point a, ∵ AB = AC, ∵ be = CE = 12bc = 16. According to Pythagorean theorem, AE = AB2 − be2 = 202 − 162 = 12. In RT △ ade, de = ad2 − AE2 = 132 − 122 = 5. When point D is on the left side of AE (as shown in the figure), BD = be-de = 16-5 = 11; when point D is on the right side of AE, BD = be + de = 16 + 5 = 21. Therefore, point D is 5 units from the midpoint of BC

In the triangle ABC, ab = AC, ab = 17, BC = 16, find the midline ad on the side of BC

AB = AC, triangle ABC is isosceles triangle, the middle line on the bottom BC is also high, using Pythagorean theorem, the middle line = √ 17 ^ 2 - (16 / 2) ^ 2 = 15

In △ ABC, ∠ BAC = 90 °, BD bisection ∠ ABC, AE ⊥ BC in E. verification: AF = ad

It is proved that: ∵ - BAC = 90 °, ∵ - ADF = 90 ° - abd. ∵ AE ⊥ BC is equal to e, ∵ - AFD = - BFE = 90 ° - DBC. ∵ BD is equal to ABC, ∵ abd = - DBC. ∵ AFD = - ADF. ≁ AF = ad

As shown in the figure, ∠ ACB = 90 ° in △ ABC, ad bisects ∠ BAC, de ⊥ AB in E

It is proved that: ∵ de ⊥ AB, ∵ AED = 90 ° = ∠ ACB, and ∵ ad bisecting ∵ BAC, ∵ DAE = ∠ DAC, ∵ ad = ad, ≌ ACD, ≌ AE = AC, ∵ ad bisecting ≁ BAC, ≁ ad ⊥ CE, that is, the straight line ad is the vertical bisector of the line CE

As shown in the figure, in the known equilateral triangle ABC, AE = CD, and AF = half BF, we prove BF ⊥ CF

Make points P and G on BF and FD respectively, make AF = FP = FG, connect PG and BG
∵ in equilateral △ ABC,
∴CE=AC-AE=BC-CD=BD
In △ abd and △ BCE
And ∠ abd = ∠ BCE
∴△ABD≌△BCE（SAS）
∴AD=BE,∠ADB=∠BEC,∠BAD=∠CBE
∵AD=BE
AF+FD=BF+FE
AF+FD=2AF+FE
FD=AF+FE
FD=FG+FE
And ∵ FD = FG + GD
∴GD=FE
In △ BDG and △ CEF
BD=CE,∠BDG=∠CEF,GD=FE
∴△BDG≌△CEF（SAS）
∴∠BGD=∠CFE
∵∠ABC=∠ABE+∠CBE=∠ABE+∠AFB=∠AFE=60°
∴∠AFE=PFG=60°,∵PF=FG
The ∧ PFG is an equilateral triangle
∴PF=PG
∵AF=BF/2,AF=PF
∴PF=PB=PG
Ψ△ BGF is a right triangle
∴∠BGD=∠CFE=90°
⊥ CF ⊥ be
(don't copy it. If you understand, write it yourself.)
(if you copy, don't kill you ~!)

As shown in the figure, De is the points on the sides BC and AC of △ ABC, and BD = CE, connecting be and ad, which intersect with point F. find the degree of ∠ AFE
ABC is an equilateral triangle

∠AFE=60°