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As shown in the figure, point I is the inner part of triangle ABC, AI intersects BC at point E, and intersection triangle ABC circumscribes circle at point D. prove (1dB = di (2id square = de * ad)
Prove: (1) point I is the heart of ABC
∴∠CBI=∠ABI;∠CBD=∠CAD=∠BAI.
In other words, DBI = DIB
∴DB=DI.
(2)∵∠CBD=∠CAD=∠BAD;∠BDE=∠ADB.
∴⊿BDE∽⊿ADB,BD/AD=DE/DB,BD*DB=DE*AD.
That is: ID & # 178; = de * ad
Point I is the inner part of the triangle, the extension line of segment AI intersects triangle ABC circumscribed at point D, and the edge of intersection BC is point E. calculate id = BD
Certification:
∵∠ bid = ∠ IBA + ∠ Bai (the outer angle is equal to the sum of two non adjacent inner angles)
∵ I is the heart, the intersection of the bisectors,
The Bi bisection ∠ B, AI bisection ∠ a,
∴∠BID=(∠A+∠B)/2
∵∠ IBD = ∠ IBE + ∠ EBD, ∠ EBD = ∠ A / 2 (equal to the circumference of arc circle)
∴∠IBD=∠BID
The ∧ DBI is an isosceles triangle,
∴ID=BD.
If id = 4, ad = 8, then de =?
If id = 4, ad = 8, then de =?
DE=2
Id = 4, ad = 8, that is, I is the midpoint of AD. the line between the center of the outer circle and I is perpendicular to AD. only when this triangle is an equilateral triangle can it meet the meaning of the problem. So de = 2
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