# Is it safe to assume that the altitude of a triangle always cuts the base in half

While solving different questions , I realized that whenever I constructed an altitude it always bisected the base in half. From what I deduced from Wikipedia is that this is only true if the triangle is either isosceles or a right triangle. What I really want to know is when I am given a question sometimes the type of triangle is not specified however I am still required to draw altitudes and assume they cut the emerging angle in half and also cut the other side in half. As shown in the following figures in red.Is there any safe way to make sure that the altitude i am drawing will definitely cut the base in half ? or do I have to first make a wise estimate of the type of triangle before constructing an altitude.

• Try drawing the altitude(s) starting at vertex B in the left-hand picture. – Dan Ramras Jun 29 '12 at 6:39
• Most triangles are not isosceles, so if one assumes it one will be mostly wrong. – André Nicolas Jun 29 '12 at 7:06
• The altitude will also bisect the opposite side of an equilateral triangle – user179163 Sep 27 '14 at 12:32

Firstly, your statement : "whenever I constructed an altitude it always bisected the base in half. From what I deduced from Wikipedia is that this is only true if the triangle is either isosceles or a right triangle" is not fully correct. An altitude from a vertex bisects the opposite base if and only if the two sides emerging from that particular vertex are equal(not necessary in a right angle triangle).Therefore, you need to specify this condition before assuming that the altitude cuts the opposite base in half.

• "An altitude from a vertex bisects the opposite base if and only if the two sides emerging from that particular vertex are equal(not necessary in a right angle triangle)" That definitely makes things a lot clearer – Rajeshwar Jun 29 '12 at 7:03

The altitude will cut the base in half only with an isoceles triangle. Take your second drawing. If CD is an altitude and BD is congruent to AD, then CDA is congruent to CDB by SAS. This means that AC and BC must be congruent.

Drawing an altitude from a right angle of a right triangle to the hypoteneuse will however divide the triangle into 2 similar triangles if that is of some use.

• One extra condition is needed: That altitude must be from the vertex common to those equal sides(in isosceles triangle). – Aang Jun 29 '12 at 6:43
• Okay so two questions : 1- So this is only true for Isosceles 2-In figure-A if AB=8 then what would AD. be isnt it cutting it in half? – Rajeshwar Jun 29 '12 at 6:46
• "Drawing an altitude from a right angle of a right triangle to the hypoteneuse will however divide the triangle into 2 similar triangles if that is of some use." So does it cut the base in two equal parts? – Rajeshwar Jun 29 '12 at 6:48
• Similarity is : $EBD \approx AED$,therefore,$AD$ is not related to $BD$ but $ED$. – Aang Jun 29 '12 at 6:55
• @Rajeshwar In your first picture, you have $ABE\sim AED\sim EBD$. If I have this right, that means $\frac{AB}{AE}=\frac{AE}{AD}$, or $(AB)(AD)=AE^2$ – Mike Jun 29 '12 at 7:06