Proof: as shown in Figure 9, in the triangle ABC, ab = AC,

Proof: as shown in Figure 9, in the triangle ABC, ab = AC,


∵AB=AC,BD=CD,
The ∧ ABC and ∧ ade are isosceles triangles
∵∠BAD=40°
∴∠DAE=40°
∴∠ADE= 12(180°-∠DAE)= 12(180°-40°)=70°
And ∵ △ ABC is an isosceles triangle
∴AD⊥CD
∴∠CDE=90°-∠ADE=90°-70°=20°.
So the answer is 20 degrees



The bisectors BP and CP of angle ABC intersect at point P. is the distance from point P to AB, BC and AC equal? Explain the reason
In a hurry


How about ABC 2 bisectors?
It should be △ ABC, right?
Of course, wait
Because BP is the bisector of ∠ ABC, the distance from point P to Ba and BC is equal
And CP is the bisector of ∠ ACB, so the distance from point P to BC and AC is equal
So the distance from point P to AB, BC and AC is equal



As shown in the figure, △ abd and △ ace are equilateral triangles, and ab


【BE=CD】
prove:
∵ abd and ace are regular triangles
∴AB=AD,AE=AC,∠BAD=∠CAE=60°
∴∠BAD+∠DAE=∠CAE+∠DAE
That is, BAE = DAC
∴△BAE≌△DAC(SAS)
∴BE=CD