One: isosceles right angle △ABC, angle A is a right angle, the bisector of angle B intersects AC at D, and the extension line of BD intersects C at E. It is proved that BD=2CE (live: Pythagorean theorem and higher-level theorem cannot be used, only some transformations of angle and the relationship between lines can be used! )
Two: AD is the height on the hypotenuse of a right triangle, the bisector BP of angle B intersects AC at P, and AD is at M, so that AQ is perpendicular to BP at Q, K is a point above AD, and AK=DK, which proves that QK⊥AD (as above).
Simple: the key is the auxiliary line! ! !
The first line: extend the extension line from CE to AB to F. From the bisector of the angle, the angle BAD= angle CAD, CE is perpendicular to BE, so the angle BEF= angle BEC=90 degrees, and the side of BE is * * *, so it can be proved that △BCE is congruent with the triangle BFE, so there is CE=FE, so CF=2CE, and angle C = 90, ABE=.
The second way: extending the intersection of AQ and BC to F, it is easy to prove that triangle ABQ is all equal to FBQ, so AQ=FQ, AK=DK, so QK is parallel to BC, and since AD is perpendicular to BC, QK is perpendicular to AD.
It should be the first grade, because I didn't learn Pythagorean theorem at that time, but I learned congruence.
Look at the picture below and read the related words. The number of intersections where ten straight lines intersect the most is (
)
Two straight lines intersect,
Three straight lines intersect,
Four straight lines intersect,
Up to 1 intersections.
Three intersections at most.
Six intersections at most.