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Let's assume a right triangle where ∠BAC =90° and a perpendicular has been drawn to hypotenuse "BC" from point "A". And this perpendicular intersects hypotenuse at point "D" such that ∠ADB and ∠ADC equal to 90°. Does this perpendicular bisects the angle ∠BAC? One video lecture says that this perpendicular does not bisect ∠BAC, i.e ∠BAD and ∠CAD are not equal, and they are not 45°. The link of this video lecture is given below:

https://m.youtube.com/watch?v=zQXAtIz_xLA&itct=CBkQpDAYACITCPmVw7Wu4tACFUvBnAod3DgOj1IkbGluZXMgYW5kIGFuZ2xlcyBnZW9tZXRyeSBwYXJ0IDcgc3Nj&client=mv-google&hl=en&gl=US

Let's assume another figure where an equilateral triangle has been drawn inside a circle. Each angle Angle of that equilateral triangle is 60°. Central angle∠AOC drawn on the arc AC is twice the inscribed angle ∠ABC drawn on that same arch AC. Since ∠ABC equals to 60°, ∠AOC is 120°. A perpendicular has been drawn to line AC, bisecting the angle ∠AOC into two equal half, i.e ∠AOD and ∠COD are equal to 60°. And OD is perpendicular to AC. (These lines about equilateral triangle drawn inside a circle have been taken from a video lecture made by Khan Academy link of which has been given below).

https://m.youtube.com/watch?v=UmiZK6Hgm6c&itct=CBYQpDAYACITCJGtoZiu4tACFa-snAodQmoDrFImYXJlYSBvZiBpbnNjcmliZWQgZXF1aWxhdGVyYWwgdHJpYW5nbGU%3D&client=mv-google&gl=US&hl=en

In case of 1, perpendicular drawn from point A to hypotenuse "BC" doesn't bisect the angle BAC, and in case of 2, perpendicular drawn from point "O" to line AC bisects ∠AOC. Please, let me know when a perpendicular drawn through an angle to a line of a triangle bisects that angle and when it doesn't.

user64814
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  • Hint: If AD does bisect BAC, then the triangles BAD and CAD have a common side and two pairs of equal angles. What can you conclude about angles B and C? –  Dec 07 '16 at 15:41
  • I'm afraid to say that I cannot understand what you have said. If angles BAD and CAD would be equal, both of them would be 45°, and hence angles B and C also would be 45°. If B and C would be equal to each other, their opposite lines also would be equal. Opposite line of angle B is AC, and in case of C it is AB. It is possible only when a right triangle is also an isosceles triangle. Am I right? Is it what you wanted to say? Otherwise, perpendicular drawn through right angle of a triangle to its hypotnuse cannot bisect that right angle. Please correct me if I am wrong. – user64814 Dec 07 '16 at 16:10
  • Please tell how I would know that perpendicular drawn through angle AOC bisects the angle AOC discussed in case of 2? – user64814 Dec 07 '16 at 16:13
  • Yes, you are right. For the second case, use the fact that the triangle AOC is isosceles to compare angles in triangles AOD and COD. –  Dec 07 '16 at 16:30
  • In the second video lecture, first the lecturer informs that perpenficular bisects the angle and then he says that triangles AOD and COD are isosceles. So first I need to know how we reach to a decision that AOC is divided by perpendicular OD? – user64814 Dec 07 '16 at 16:45

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