Triangle ABC is a right triangle with right angle at A. Suppose AX is an altitude of the triangle, AY is an angle bisector of the triangle, and AZ is a median of the triangle, and angle XAY = 13 degrees. If X is on BY, then what is the measure of angle ZAC?

bbelt711 Aug 3, 2017

#1**+3 **

Since AY bisects the 90° angle... m∠BAY = 90° / 2 = 45°

So... m∠BAX = 45° - 13° = 32°

And since there are 180° in triangle BAX... m∠ABX = 180° - 32° - 90° = 58°

And since there are 180° in triangle ABC... m∠BCA = 180° - 58° - 90° = 32°

Now..to show why triangle ACZ is isocelese....

Draw a line from Z to side BA that is parallel to AC.

Draw a line from Z to side AC that is parallel to BA.

Since Z is the midpoint of BC, BZ = ZC .

So we can be sure that triangle BJZ is congruent to triangle ZHC from the AAS rule.

And... AH = JZ = HC

So.. we can be sure that triangle AHZ is congruent to triangle CHZ by the SAS rule.

Therefore... m∠ACZ = m∠ZAC = 32°

hectictar Aug 4, 2017

#1**+3 **

Best Answer

Since AY bisects the 90° angle... m∠BAY = 90° / 2 = 45°

So... m∠BAX = 45° - 13° = 32°

And since there are 180° in triangle BAX... m∠ABX = 180° - 32° - 90° = 58°

And since there are 180° in triangle ABC... m∠BCA = 180° - 58° - 90° = 32°

Now..to show why triangle ACZ is isocelese....

Draw a line from Z to side BA that is parallel to AC.

Draw a line from Z to side AC that is parallel to BA.

Since Z is the midpoint of BC, BZ = ZC .

So we can be sure that triangle BJZ is congruent to triangle ZHC from the AAS rule.

And... AH = JZ = HC

So.. we can be sure that triangle AHZ is congruent to triangle CHZ by the SAS rule.

Therefore... m∠ACZ = m∠ZAC = 32°

hectictar Aug 4, 2017

#2**+2 **

Excellent thinking, hectictar....!!!!!....this one was a little tricky, for sure....!!!!!

Here's one more thought......dropping a perpendicular from Z to H on AC means that ZH is parallel to BA....and whenevever a segment is drawn parallel to a base, it splits the sides of the triangle into equal ratios....that is.....CZ / BZ = CH / AH.....but since BZ = CZ [ because Z is the midpoint of BC ], then AH = CH

Then by SAS, triangle CHZ is congruent to triangle AHZ.....and since you found that BCA = 32° = ZCH = ZAH = ZAC

Obviously.....dropping that perpendicular as you did was the key to the whole thing.....not bad for a 'Bama fan....LOL!!!!

CPhill Aug 4, 2017