In the triangle ABC, BC = 2Ab, ∠ ABC = 2 ∠ C, BD = CD

In the triangle ABC, BC = 2Ab, ∠ ABC = 2 ∠ C, BD = CD

prove:
If the angle bisector BD passes through B and intersects with AC at D, then
∵∠ABC=2∠ACB,BC=2AB,
∴∠DBC=∠ABC,
The ∧ DBC is an isosceles triangle,
Let d be de ⊥ BC over e, then E is also the midpoint of BC, that is EB = EC = (1 / 2) BC = ab,
∵DB=DB,∠DBC=∠DBA,
∴△DBE≌△DBA,
∴∠DAB=∠DEB=90°
That is, BAC = 90 °,
The ABC is a right triangle,
Get it!

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