The isosceles triangle ABC AB = AC, the angle BAC = 45 degrees, ad vertical BC to D, CE vertical AB to e, ad and EC intersect h, prove that ah is equal to BC

The isosceles triangle ABC AB = AC, the angle BAC = 45 degrees, ad vertical BC to D, CE vertical AB to e, ad and EC intersect h, prove that ah is equal to BC

prove:
∵CE⊥AB,∠BAC=45°
∴∠ACE=45°
∵AD⊥BC
∴∠EAH=90°- ∠B=∠ECB
∴Rt△AEH≌Rt△CEB（HL）
∴AH=CB
[this question has nothing to do with ab = AC]

H is the vertical center of △ ABC, ah ⊥ BC in D, BH ⊥ AC in E, CH ⊥ AB in F. the proof is DH / DA + eh / EB + FH / FC = 1

If s △ HBC / s △ ABC = & # 189; BC · DH / (& # 189; BC · DA) = DH / DA, then:
DH/DA+EH/EB+FH/FC
=S△HBC/S△ABC+S△AHC/S△ABC+S△AHB/S△ABC
=(S△HBC+S△AHC+S△AHB)/S△ABC
=S△ABC/S△ABC
=1

As shown in the figure, △ ABC, D is the midpoint of AB, AC = 12, BC = 5, CD = 13 / 2

cosA=(AD^2+AC^2-CD^2)/2AD*AC
cosA=(AB^2+AC^2-BC^2)/2AB*AC
∵ D is the midpoint of ab
∴AD=AB/2
∴AB=2AD
∴(AD^2+AC^2-CD^2)/2AD*AC=((2AD)^2+AC^2-BC^2)/2(2AD)*AC
(AD^2+144-169/4) / 2AD*AC =(4AD^2+144-25)/4AD*AC
2(AD^2+144-169/4)=4AD^2+144-25
4(AD^2+144-169/4)=2(4AD^2+144-25)
4AD^2+576-169=8AD^2+288-50
4AD^2=169
AD^2=169/4
AD=13/2
AB=13
∵AB^2=169
AC^2=144
BC^2=25
AC^2+BC^2=144+25=169=AB^2
The ABC is a right triangle

In a triangle, D is the midpoint of AB side, AC = 12, BC = 5, CD = 6.5. Prove that ABC is a right triangle

According to Pythagorean, ab = 13D is the midpoint of AB, ad = DB = 6.5, CD = 6.5, so ad = DB = cdacd and BCD are isosceles triangle angle a = angle DCA angle B = angle DCB. According to the law of exterior angle of triangle, angle CDB = angle a + angle DCA angle CDA = angle B + angle DCB and angle CDB + angle CDA = angle a + angle DCA + angle B + angle DCB = 2 (angle a + angle b) = 180