As shown in the figure, in △ ABC, ab = AC, ad ⊥ BC is at point D, de ∥ AB intersects AC at point E. is △ ade an isosceles triangle? Why?

As shown in the figure, in △ ABC, ab = AC, ad ⊥ BC is at point D, de ∥ AB intersects AC at point E. is △ ade an isosceles triangle? Why?

∵AB=AC，
∴∠B=∠C，
∵ de ∫ AB intersects AC at point E,
∴∠B=∠EDC，
∴∠EDC=∠C，
∴ED=EC，
∵AD⊥BC，
∴AE=CE=DE．
The △ ade is an isosceles triangle

As shown in the figure, in the isosceles triangle ABC, ab = AC, D is a point on the extension line of BC, Confirmation: Ad 2 = AB 2 + BD · DC

Make AE ⊥ BC through vertex a of triangle, and intersect BC with E,
According to the Pythagorean theorem,
AB＾2＝AE＾2＋BE＾2,
AD＾2＝AE＾2＋ED＾2
AB^2-AD^2＝AE＾2＋BE＾2－AE＾2－ED＾2
＝BE＾2－ED^2
=(BE+ED)(BE-ED)
=BD*(EC-ED)
=BD*DC

As shown in the graph, in △ ABC, ab = AC, point D is on AC, and BD = BC = ad (1), which isosceles triangles are there in the graph? (2) What are the equal angles in the graph and why? (3) Calculate the degree of each angle of △ ABC; (4) What conclusions can you draw from the above answers? Please summarize them (write at least two)

1. The △ ABC, △ BCD, △ ADB is the isosceles triangle, 2, and ＜ ABC is the isoscetriangle, 2, ∠ ABC = ∠ ACB, ∠ a = ∠ abd, ∠ BCD = ∠ BDC = ∠ ACB = ∠ ACB = ∠ Abc3, the ∵ BDC = ∠ ACB = ∠ ABC = (a + (a) + ABC = 2 ∠ ∵ a ∵ a ∵ a ∵ ABC ∵ ABC ∵ ABC

As shown in the figure, in △ ABC, ad is the height on BC edge, e and F are the midpoint on AB and AC respectively. Is △ def similar to △ ABC

Δ DEF is similar to △ ABC
∵ E and F are the midpoint of AB and AC respectively
∴EF‖BC
∴△AEF∽△ABC
Let EF and ad intersect o, then Ao = do
∵AD⊥BC
∴AD⊥EF
∴AE=DE,AF=DF
∵EF=EF
∴△AEF≌△DEF
∴,△DEF∽△ABC

The angle bisector of the ad triangle ABC is known, and ab = AC + DC

Take a point m on AB to make am = AC and connect MD, then
Δ amd congruent △ ACD, ν MD = CD, ∠ C = ∠ AMD
∵AB=AC+DC,∴MD=BM,∴∠B=∠MDB
Amd = ∠ B + ∠ MDB
ν angle c = 2 angle B

In the triangle ABC, ad is the bisector of angle A. It is proved that the square of ad = AB * ac-bd * DC

Take a point m on AC to make the angle ADM = angle ABC, so the triangle abd is similar to the triangle ADM, so AB / ad = ad / am, i.e. ad ^ 2 = AB * am. We only need to prove that ab * am = AB * ac-bd * DC, that is, ab * MC = BD * DC. From the graph, we can see that the triangle CMD is similar to the triangle CDA, so cm * CD = CD * AC is subdivided by the triangle angle

It is known that in the triangle ABC, ad is the bisector of ∠ A. It is proved that ab ratio AC = BD ratio DC

It is proved that if the parallel line of AB is made through point C and the extension line of ad is at point E, then AC = EC
From triangle abd similar to triangle ECD, AB / CE = BD / DC
Therefore, AB / AC = BD / DC

As shown in the figure, △ ABC, ab = BC = AC, ∠ B = ∠ C = 60 °, BD = CE, where ad and be intersect at point P, then the degree of ∠ ape is () A. 45° B. 55° C. 75° D. 60°

In △ abd and △ BCE,
AB＝BC
∠ABD＝∠BCE
BD＝CE ，
∴△ABD≌△BCE（SAS），
∴∠BAD=∠CBE，
∵∠APE=∠ABE+∠BAD，∠ABE+∠CBE=60°，
∴∠APE=∠ABC=60°．
Therefore, D

As shown in the figure, △ ABC, ab = BC = AC, ∠ B = ∠ C = 60 °, BD = CE, where ad and be intersect at point P, then the degree of ∠ ape is () A. 45° B. 55° C. 75° D. 60°

In △ abd and △ BCE,
AB＝BC
∠ABD＝∠BCE
BD＝CE ，
∴△ABD≌△BCE（SAS），
∴∠BAD=∠CBE，
∵∠APE=∠ABE+∠BAD，∠ABE+∠CBE=60°，
∴∠APE=∠ABC=60°．
Therefore, D

As shown in the figure, △ ABC, ab = BC = AC, ∠ B = ∠ C = 60 °, BD = CE, where ad and be intersect at point P, then the degree of ∠ ape is () A. 45° B. 55° C. 75° D. 60°

In △ abd and △ BCE,
AB＝BC
∠ABD＝∠BCE
BD＝CE ，
∴△ABD≌△BCE（SAS），
∴∠BAD=∠CBE，
∵∠APE=∠ABE+∠BAD，∠ABE+∠CBE=60°，
∴∠APE=∠ABC=60°．
Therefore, D