AD is an altitude of ΔABC with D on BC and DB
I just really need to know how to get started... I can't quite seem to get my head around it quite yet... I drew the picture, but then I got stuck....
2007-01-30 17:50:39 · 3 answers · asked by yogastar02 2 in Science & Mathematics ➔ Mathematics
Add to the picture point B' on BC, in between D and C, and so that distance DB = DB', and draw AB'. Since AD is an altitude, angle ADB = angle ADB' = 90°, AD = AD, and DB = DB', so triangles ADB and ADB' are congruent by SAS, and therefore angle BAD = angle B'AD.
Now angle CAD = angle CAB' + angle B'AD, and since angle CAB' > 0, angle CAD must be greater than angle BAD.
2007-01-30 18:08:42 · answer #1 · answered by Philo 7 · 0⤊ 0⤋
In any triangle, a bigger side is always across from a bigger angle. So if you know that angle BAD is opposite from side BD, and CD is larger than BD, then angle CAD must be larger because it is across from a longer side.
2007-01-30 17:55:15 · answer #2 · answered by T F 4 · 0⤊ 0⤋
tan(BAD) = DB/AD
tan(CAD) = DC/AD
DB = ADtan(BAD)
DC = ADtan(CAD)
DB < DC, so
ADtan(BAD) < ADtan(CAD)
tan(BAD) < tan(CAD)
BAD < CAD
2007-01-30 18:50:25 · answer #3 · answered by Helmut 7 · 0⤊ 0⤋