Triangle ABC, ad is perpendicular to BC, BC intersects at D, ad has a point E, connect be and EC, prove be-ec > bd-dc You can use the difference between the two sides less than the third side to do, Pythagorean theorem has not been learned

Triangle ABC, ad is perpendicular to BC, BC intersects at D, ad has a point E, connect be and EC, prove be-ec > bd-dc You can use the difference between the two sides less than the third side to do, Pythagorean theorem has not been learned

According to Pythagorean theorem be BD = EC DC
That is be EC = BD DC
Factorization (be + EC) (be-ec) = (BD + DC) (bd-dc)
(be + EC) / (BD + DC) is always greater than 1, so I calculate be-ec < bd-dc
I don't know if I miscalculated and didn't match the original question

In the triangle ABC, D and E are connected with BD and Ce on AB and AC respectively, and the angle BCE = angle CBE = angle A and be = DC are proved

There are two problems in the title: D and E are on AB and AC respectively? BCE = CBE?

As shown in Figure 2, take sides AB and AC of triangle ABC as sides, make equilateral triangle Abe and equilateral triangle ACF on the outside of triangle ABC respectively, and CE intersects BF o
1. Find the relationship between EC and BF
2. Calculate the degree of angle EOB

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